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CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 4 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Indefinite Integrals
Quiz 4
If
∫
x
log
(
1
+
1
x
)
d
x
=
f
(
x
)
log
(
x
+
1
)
+
g
(
x
)
.
x
2
+
L
x
+
c
, then
Report Question
0%
f
(
x
)
=
x
2
2
−
1
2
,
g
(
x
)
=
1
2
log
x
,
L
=
1
0%
f
(
x
)
=
x
2
2
+
1
2
,
g
(
x
)
=
−
1
2
log
x
,
L
=
1
2
0%
f
(
x
)
=
x
2
2
−
1
2
,
g
(
x
)
=
−
1
2
log
x
,
L
=
1
2
0%
f
(
x
)
=
x
2
2
−
1
2
,
g
(
x
)
=
1
2
log
x
,
L
=
−
1
2
Explanation
∫
x
log
(
1
+
1
x
)
d
x
.
=
log
(
1
+
1
x
)
⋅
x
2
2
−
∫
(
d
d
x
log
(
1
+
1
x
)
∫
x
d
x
)
⋅
d
x
=
log
(
1
+
1
x
)
⋅
x
2
2
−
∫
x
(
x
+
1
x
)
⋅
−
1
x
x
d
x
.
=
log
(
1
+
1
x
)
⋅
x
2
2
+
∫
x
x
+
1
⋅
d
x
.
=
log
(
1
+
1
x
)
⋅
x
2
2
+
x
−
log
(
x
+
1
)
.
=
[
log
(
1
+
x
)
]
x
2
2
∫
x
2
(
log
x
)
3
d
x
=
Report Question
0%
x
3
[
(
l
o
g
x
)
3
3
−
(
l
o
g
x
)
2
3
+
2
(
l
o
g
x
)
9
−
2
27
]
+
c
0%
x
3
[
(
log
x
)
3
3
+
(
log
x
)
2
3
+
2
(
log
x
)
9
+
2
27
]
+
c
0%
x
3
[
(
log
x
)
3
3
−
(
log
x
)
2
27
]
+
c
0%
x
3
[
(
log
x
)
3
3
+
(
log
x
)
2
3
+
2
(
log
x
)
9
−
2
27
]
+
c
Explanation
∫
x
2
(
l
o
g
x
)
3
d
x
=
(
l
o
g
x
)
3
∫
x
2
d
x
=
−
∫
(
d
d
x
(
l
o
g
x
)
3
.
∫
x
2
.
d
x
)
.
d
x
=
(
l
o
g
x
)
3
x
3
3
∫
3
(
l
o
g
x
)
2
−
1
/
x
.
x
2
/
3
d
x
(
l
o
g
x
)
3
x
3
3
−
∫
(
l
o
g
x
)
2
.
d
x
∫
x
2
(
l
o
g
x
)
2
=
(
l
o
g
x
)
2
x
3
3
−
∫
1
/
2
l
o
g
x
x
.
x
2
/
3
=
(
l
o
g
x
)
2
−
x
3
/
3
−
1
/
6
∫
l
o
g
x
.
x
2
∫
l
o
g
x
.
x
2
=
l
o
g
x
.
x
3
/
3
−
∫
1
/
x
x
2
/
3
d
x
=
l
o
g
x
−
x
3
/
3
−
x
3
/
q
+
c
=>
x
3
[
(
l
o
g
x
)
3
−
(
l
o
g
x
)
2
+
2
9
l
o
g
x
−
2
27
]
+
c
∫
x
s
i
n
2
x
d
x
Report Question
0%
x
2
4
−
cos
2
x
3
+
1
8
s
i
n
2
x
+
c
0%
x
2
4
−
x
s
i
n
2
x
4
−
1
8
c
o
s
2
x
+
c
0%
x
2
4
+
x
s
i
n
2
x
4
+
1
8
c
o
s
2
x
+
c
0%
−
x
2
4
−
c
o
s
2
x
3
+
1
8
c
o
s
2
x
+
c
Explanation
∫
x
s
i
n
2
x
d
x
.
=
x
∫
s
i
n
2
x
.
d
x
−
∫
(
∫
s
i
n
2
x
.
d
x
)
d
x
c
o
s
2
x
=
1
−
2
s
i
n
2
x
2
s
i
n
2
x
=
1
−
c
o
s
2
x
2
x
∫
(
1
−
c
o
s
2
x
2
)
d
x
−
∫
(
∫
(
1
−
c
o
s
2
x
2
)
.
d
x
)
d
x
.
=
x
2
[
x
−
1
/
2
s
i
n
2
x
]
−
∫
[
x
2
−
1
/
2
s
i
n
2
x
]
d
x
=
x
2
2
−
x
4
s
i
n
2
x
−
∫
[
x
2
4
+
1
4
c
o
s
2
x
]
=
x
2
4
−
x
s
i
n
2
x
4
−
1
8
c
o
s
2
x
+
c
∫
x
1
+
cos
x
d
x
=
Report Question
0%
x
tan
x
2
−
2
log
|
sec
x
/
2
|
+
c
0%
−
x
tan
x
/
2
−
1
2
log
|
sec
x
/
2
|
+
c
0%
x
tan
x
/
2
+
1
2
log
|
sec
x
/
2
|
+
c
0%
x
cot
x
/
2
−
1
2
log
|
csc
x
/
2
|
+
c
Explanation
∫
x
1
+
c
o
s
x
d
x
.
=
x
∫
1
1
+
c
o
s
x
d
x
−
∫
1
(
∫
1
1
+
c
o
s
x
d
x
)
d
x
∫
1
1
+
c
o
s
x
d
x
=
t
a
n
x
2
x
t
a
n
x
2
−
∫
t
a
n
x
2
.
d
x
.
x
t
a
n
x
2
+
2
l
o
g
|
c
o
s
x
|
+
c
=
x
t
a
n
x
2
−
2
l
o
g
|
s
e
c
x
2
|
+
c
Area bounded by
y
=
{
x
}
,
{
.
}
is fractional part of function and
x
=
±
1
is in sq. units
Report Question
0%
1
0%
2
0%
3
0%
4
Explanation
Area
=
2
∫
1
0
x
d
x
=
2
[
x
2
2
]
1
0
=
2.
1
2
=
1
∫
c
o
s
(
log
x
)
d
x
=
Report Question
0%
x
[
c
o
s
(
log
x
)
−
s
i
n
(
log
x
)
]
+
c
0%
x
2
[
c
o
s
(
log
x
)
−
s
i
n
(
log
x
)
]
+
c
0%
log
x
2
[
c
o
s
x
+
s
i
n
x
]
+
c
0%
x
2
[
c
o
s
(
log
x
)
+
s
i
n
(
log
x
)
]
+
c
Explanation
∫
c
o
s
(
l
o
g
x
)
1
d
x
x
c
o
s
(
l
o
g
x
)
+
∫
(
s
i
n
(
l
o
g
x
)
x
⋅
x
)
d
x
=
x
c
o
s
(
l
o
g
x
)
+
∫
s
i
n
(
l
o
g
x
)
d
x
∫
s
i
n
(
l
o
g
x
)
d
x
=
x
s
i
n
(
l
o
g
x
)
+
∫
c
o
s
(
l
o
g
x
)
d
x
∫
c
o
s
(
l
o
g
x
)
d
x
=
x
c
o
s
(
l
o
g
x
)
+
x
s
i
n
(
l
o
g
x
)
−
∫
c
o
s
(
l
o
g
x
)
d
x
=
∫
c
o
s
(
l
o
g
x
)
d
x
=
x
2
[
c
o
s
(
l
o
g
x
)
+
s
i
n
(
l
o
g
x
)
]
+
c
.
∫
log
(
log
x
)
x
d
x
=
Report Question
0%
log
x
[
log
(
log
x
)
−
1
]
+
c
0%
x
[
log
(
log
x
)
+
1
]
+
c
0%
log
x
[
log
(
log
x
)
+
1
]
+
c
0%
x
[
log
(
log
x
)
−
1
]
+
c
Explanation
∫
l
o
g
(
l
o
g
x
)
x
d
x
=
l
o
g
(
l
o
g
x
)
(
l
o
g
x
)
−
∫
(
1
l
o
g
x
1
x
l
o
g
x
)
d
x
=
l
o
g
(
l
o
g
x
)
[
l
o
g
x
]
−
l
o
g
x
+
c
=
l
o
g
x
[
l
o
g
(
l
o
g
x
)
−
1
]
+
c
∫
2
m
x
.3
n
x
d
x
when
m
,
n
ϵ
N
is equal to:
Report Question
0%
2
m
x
+
3
n
x
m
ln
2
+
n
ln
3
+
c
0%
e
(
m
x
ln
2
+
n
x
ln
3
)
m
ln
2
+
n
ln
3
+
c
0%
2
m
x
.3
n
x
ln
(
2
m
⋅
3
n
)
+
c
0%
None of these
Explanation
Let,
I
=
∫
2
m
x
.3
n
x
d
x
Now put
2
m
=
a
,
3
n
=
b
⇒
I
=
∫
(
a
x
.
b
x
)
d
x
Now using by parts, by considering
a
x
as 1st function and
b
x
as 2nd function
⇒
I
=
a
x
b
x
ln
b
−
∫
a
x
ln
a
.
b
x
ln
b
d
x
+
k
⇒
I
=
a
x
b
x
ln
b
−
ln
a
ln
b
∫
a
x
b
x
d
x
+
k
⇒
I
=
a
x
b
x
ln
b
−
ln
a
ln
b
I
+
k
⇒
I
=
a
x
b
x
ln
a
+
ln
b
+
k
ln
b
ln
a
+
ln
b
⇒
I
=
a
x
b
x
ln
a
+
ln
b
+
c
, where
c
=
k
ln
b
ln
a
+
ln
b
∈
R
Hence option B,C are correct choice
lim
is equal to
Report Question
0%
2
0%
\displaystyle -\frac{1}{2}
0%
\displaystyle \frac{1}{2}
0%
1
Explanation
\displaystyle \lim _{ x\to \infty }{ \int _{ 0 }^{ x } xe^{ t^{ 2 }-x^{ 2 } }dt } =\lim _{ x\to \infty }{ \int _{ 0 }^{ x } \dfrac { xe^{ t^{ 2 } } }{ { e }^{ { x }^{ 2 } } } dt } =\lim _{ x\to \infty }{ \int _{ 0 }^{ x } \dfrac { e^{ t^{ 2 } } }{ \dfrac { { e }^{ { x }^{ 2 } } }{ x } } dt }
\displaystyle \lim _{ x\to \infty }{ \dfrac { { e }^{ { x }^{ 2 } } }{ \dfrac { x(2x){ e }^{ { x }^{ 2 } }-{ e }^{ { x }^{ 2 } } }{ x^{ 2 } } } } =\lim _{ x\to \infty }{ \dfrac { { e }^{ { x }^{ 2 } } }{ { e }^{ { x }^{ 2 } }\left( \dfrac { 2x^{ 2 }-1 }{ x^{ 2 } } \right) } } =\lim _{ x\to \infty }{ \dfrac { 1 }{ 2-\dfrac { 1 }{ x^{ 2 } } } } =\dfrac { 1 }{ 2 }
\displaystyle \int \sin^{-1}\left ( \frac{2x}{1+x^2} \right )dx
is equal to
Report Question
0%
\displaystyle 2(x \tan^{-1} x +\ln \left | \cos (\tan^{-1}x) \right |)+C
0%
2[(\displaystyle x\tan^{-1}{x})^2+\ln \left | \sec (\tan^{-1}x) \right |]+c
0%
2[(\displaystyle x\tan^{-1}{x})^2-\ln \left | \cos (\tan^{-1}x) \right |]+c
0%
None of these
Explanation
\displaystyle I=\int \sin^{-1}\left ( \frac{2x}{1+x^{2}} \right )dx
Let
x=\tan \theta
\implies \displaystyle dx=\sec ^{2}\theta d\theta
\therefore \displaystyle I=\int \sin^{-1}\left ( \frac{2\tan \theta }{1+\tan^{2}\theta} \right ) \sec^{2}\theta d\theta
=\displaystyle \int \sin^{-1}\left(\dfrac{2\frac{\sin \theta}{\cos \theta}}{1+\frac{\sin^{2}\theta}{\cos^{2}\theta}}\right)\sec^{2}\theta d\theta
=\displaystyle \int \sin^{-1}\left(\dfrac{2\frac{\sin \theta}{\cos \theta}}{\frac{\sin^{2}\theta+\cos^{2}\theta}{\cos^{2}\theta}}\right)\sec^{2}\theta d\theta
=\displaystyle \int \sin^{-1}(2\sin{\theta} \cos{\theta})\sec^{2}\theta d\theta
=\displaystyle \int \sin^{-1}(\sin{2\theta})\sec^{2}\theta d\theta
=\displaystyle 2\int \theta \sec ^{2}\theta \:d\theta
=\displaystyle 2(\theta \:\tan \:\theta +\ln \left | \cos \:\theta \right |)+C
=\displaystyle 2(x \tan^{-1} x +\ln \left | \cos (\tan^{-1}x) \right |)+C
Ans: A
\displaystyle \int e^{tan^{-1}x}\left[\frac{1+x+x^{2}}{1+x^{2}}\right]dx=
Report Question
0%
\displaystyle x^{2}e^{\tan^{-1}x}+c
0%
\displaystyle x e^{\tan^{-1}x}+c
0%
\displaystyle e^{\tan^{-1}x}+c
0%
\displaystyle \frac{1}{2}e^{\tan^{-1}x}+c
Explanation
Let
I=\displaystyle \int e^{\tan^{-1}x}\left[\frac{1+x+x^{2}}{1+x^{2}}\right]dx
=\displaystyle \int e^{\tan^{-1}x}\left[1+\frac {x}{1+x^2}\right]dx
=\displaystyle \int \left[e^{\tan^{-1}x}+\frac {xe^{\tan^{-1}x}}{1+x^2}\right]dx
For
f(x)=e^{\tan^{-1} x}
, above integral
I
reduces in the form of
\displaystyle \int [f(x)+x f^{'}(x)]dx
But we know that,
\dfrac{d}{dx} (xf(x)+c) = \displaystyle [f(x)+x f^{'}(x)]
\therefore \displaystyle \int [f(x)+x f^{'}(x)]dx=xf(x)+c
\therefore I = \displaystyle xe^{\tan^{-1}x}+c
\therefore I=\displaystyle \int e^{\tan^{-1}x}\left[\frac{1+x+x^{2}}{1+x^{2}}\right]dx=xe^{\tan ^{-1} x}+c
Hence, option 'B' is correct.
If
\displaystyle \int x^{2}e^{-2x}dx=e^{-2x}(ax^{2}+bx+c)+d
then
Report Question
0%
a=-\displaystyle \frac {1}{2}
0%
b=2
0%
c=-\displaystyle \frac{1}{4}
0%
d\in R
Explanation
\displaystyle \int x^{2}e^{-2x}dx=e^{-2x}(ax^{2}+bx+c)+d
Differentiating both sides, we get
x^{2}e^{-2x}=e^{-2x}(2ax+b)+(ax^{2}+bx+c)(-2e^{-2x})
x^2e^{-2x}=e^{-2x}(-2ax^{2}+2(a-b)x+b-2c)
On comparing the coefficients, we get
-2a=1,\:2(a-b)=0,\:b-2c=0
\displaystyle a=\frac{-1}{2}, b=-\frac{1}{2},\:c=-\frac{1}{4}
Also,
d\in R
\displaystyle \int x.\frac{\ln \left ( x+\sqrt{1+x^2} \right )}{\sqrt{1+x^2}} dx
equals
Report Question
0%
\sqrt{1+x^2}\ln \left ( x+\sqrt{1+x^2} \right )-x+c
0%
\displaystyle \frac{x}{2}.\ln^2\left ( x+\sqrt{1+x^2} \right )-\displaystyle \frac{x}{\sqrt{1+x^2}}+c
0%
\displaystyle \frac{x}{2}.\ln^2\left ( x+\sqrt{1+x^2} \right )+\displaystyle \frac{x}{\sqrt{1+x^2}}+c
0%
\sqrt{1+x^2}\ln \left ( x+\sqrt{1+x^2} \right )+x+c
Let
\displaystyle \int e^{x}\left \{ f(x)-f'(x) \right \}dx=\phi (x).
Then
\int e^{x} f(x)dx
is
Report Question
0%
\displaystyle \phi (x)=e^{x}f(x)
0%
\displaystyle \phi (x)-e^{x}f(x)
0%
\displaystyle \frac{1}{2}\left \{ \phi (x)+e^{x}f(x) \right \}
0%
\displaystyle \frac{1}{2}\left \{ \phi (x)+e^{x}f'(x) \right \}
Explanation
Given
\displaystyle \int e^{x}\left \{ f(x)-f'(x) \right \}dx=\phi (x).
...(1)
Consdier,
\int e^{ x }\left\{ { f }(x)-{ f }'(x) \right\} dx
=\int e^{ x }f(x)dx-\int e^{ x }f'(x)dx
=e^{ x }f(x)-2\int e^{ x }f'(x)dx
So, by (1),
e^{ x }f(x)-2\int e^{ x }f'(x)dx=\phi (x)
\displaystyle \int e^xf(x)dx =\frac{1}{2}\left \{ \phi (x)+e^{x}f(x) \right \}
\displaystyle \int e^{\displaystyle \tan x}(\sec x-\sin x)dx
is equal to
Report Question
0%
\displaystyle e^{\displaystyle \tan x}\cos x+C
0%
\displaystyle e^{\displaystyle \tan x}\sin x+C
0%
\displaystyle -e^{\displaystyle \tan x}\cos x+C
0%
\displaystyle e^{\displaystyle \tan x}\sec x+C
Explanation
I=\displaystyle \int e^{\tan x}(\sec x-\sin x)dx
=-\displaystyle \int \sin xe^{\tan x}dx+ \int \sec xe^{\tan x}dx
=\displaystyle e^{\tan x}\cos x-\int \cos xe^{\tan x}\sec ^{2}xdx+\int \sec xe^{\tan x}dx
=\displaystyle \cos xe^{\tan x}+C
\displaystyle \int \left ( \log x \right )^{2}dx.
Report Question
0%
\displaystyle x \left [\left ( \log x \right )^{2}-2 \log x+2 \right ].
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\displaystyle x \left [\left ( \log x \right )^{2}-2 \log x+1 \right ].
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\displaystyle x \left [\left ( \log x \right )^{2}-2 \log x-2 \right ].
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\displaystyle x \left [\left ( \log x \right )^{2}+2 \log x-2 \right ].
Explanation
Integrating by parts,
\displaystyle I=x \left ( \log x \right )^{2}-2\int x.\log x.\frac{1}{x}dx
\displaystyle =x \left ( \log x \right )^{2}-2\int \log x\:dx
\displaystyle =x \left ( \log x \right )^{2}-2\left [ x\log x-x \right ]
\displaystyle =x \left [\left ( \log x \right )^{2}-2 \log x+2 \right ].
\displaystyle \int x^{2}\tan ^{-1}x\:dx.
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\displaystyle \frac{x^{3}}{3}\tan^{-1}x-\frac{1}{6}x^{2}+\frac{1}{6}\log\left ( x^{2}+1 \right ).
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\displaystyle \frac{x^{3}}{3}\tan^{-1}x+\frac{1}{6}x^{2}+\frac{1}{6}\log\left ( x^{2}+1 \right ).
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\displaystyle \frac{x^{3}}{3}\tan^{-1}x-\frac{1}{3}x^{2}+\frac{1}{6}\log\left ( x^{2}+1 \right ).
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\displaystyle \frac{x^{3}}{3}\tan^{-1}x-\frac{1}{6}x^{2}+\frac{1}{3}\log\left ( x^{2}+1 \right ).
Explanation
Integrating by parts,
\displaystyle I=\frac{x^{3}}{3}\tan^{-1}x-\frac{1}{3}\int \frac{x^{3}}{x^{2}+1}dx.
=\displaystyle \frac{x^{3}}{3}\tan^{-1}x-\frac{1}{3}\int \left ( x-\frac{x}{x^{2}+1}\right )dx
=\displaystyle \frac{x^{3}}{3}\tan^{-1}x-\frac{1}{6}x^{2}+\frac{1}{6}\log\left ( x^{2}+1 \right )+c.
\displaystyle \int\frac{\log x}{(1+x)^{3}}dx
is equal to
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\displaystyle \frac{\log x}{(1+x)^{2}}+\displaystyle \frac{1}{2}\log\frac{x}{x+1}+\frac{1}{2}\frac{1}{x+1}+c
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\displaystyle 2\frac{-\log x}{(1+x)^{2}}+\displaystyle 2\log\frac{x+1}{x}+\frac{2}{x+1}+c
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\displaystyle \frac{-2\log x}{(1+x)^{2}}+\displaystyle 2\log\frac{x}{x+1}+\frac{2}{x+1}+c
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\displaystyle \frac{-\log x}{(1+x)^{2}}+\displaystyle \frac{1}{2}\log\frac{x+1}{x}-\frac{1}{2}\frac{1}{x+1}+c
Explanation
I=\int \dfrac { \log x }{ (1+x)^{ 3 } } dx
\displaystyle =\log x\int \dfrac { 1 }{ (1+x)^{ 3 } } dx-\int \dfrac { -2 }{ x(1+x)^{ 2 } } dx
\displaystyle =\dfrac { -2\log x }{ (1+x)^{ 2 } } +2\int \dfrac { 1 }{ x(1+x)^{ 2 } } dx
Now,
\displaystyle \dfrac { 1 }{ x(1+x)^{ 2 } } =\dfrac { A }{ x } +\dfrac { B }{ (1+x) } +\dfrac { C }{ (1+x)^{ 2 } }
\Rightarrow 1=A(1+x)^{ 2 }+Bx(1+x)+Cx
When
x=-1, C=-1
When
x=0,A=1
When
x=1,B=-1
So
\displaystyle \frac { 1 }{ x(1+x)^{ 2 } } =\frac { 1 }{ x } -\frac { 1 }{ (1+x) } -\frac { 1 }{ (1+x)^{ 2 } }
\displaystyle \int { \frac { 1 }{ x(1+x)^{ 2 } } dx } =\int { \frac { 1 }{ x } dx } -\int { \frac { dx }{ (1+x) } } -\int { \frac { dx }{ (1+x)^{ 2 } } }
\displaystyle =\log { x } - \log { (x+1) } +\frac { 1 }{ (x+1) }
So,
\displaystyle I=\frac { -2\log x }{ (1+x)^{ 2 } } +2\log { x } -2\log { (x+1) } +\frac { 2 }{ (x+1) } +C
\displaystyle I=\frac { -2\log x }{ (1+x)^{ 2 } } +2\log { \frac { x }{ x+1 } } +\frac { 2 }{ (x+1) } +C
\displaystyle \int x\sec^{2}2xdx
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\displaystyle \frac{1}{4}x\tan 2x-\frac{1}{2}\log \sec 2x.
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\displaystyle \frac{1}{2}x\tan 2x+\frac{1}{4}\log \sec 2x.
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\displaystyle \frac{1}{4}x\tan 2x-\frac{1}{4}\log \sec 2x.
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\displaystyle \frac{1}{2}x\tan 2x+\frac{1}{4}\log \cos 2x.
Explanation
Let
\displaystyle I=\int { x\sec ^{ 2 }{ 2x } dx }
\displaystyle =\frac { 1 }{ 2 } x\tan { 2x } -\frac { 1 }{ 2 } \int { \tan { 2x } dx }
\displaystyle =\frac { 1 }{ 2 } x\tan { 2x } -\frac { 1 }{ 4 } \log { \sec { 2x } }
\displaystyle =\frac { 1 }{ 2 } x\tan { 2x } +\frac { 1 }{ 4 } \log { \cos { 2x } }
Hence, option 'D' is correct.
\displaystyle\int \left ( \log x \right )^{2}dx.
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\displaystyle x\left ( \log x \right )^{2}+2x \log x+2x
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\displaystyle x\left ( \log x \right )^{2}-2x \log x+2x
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\displaystyle x\left ( \log x \right )^{2}-2x \log x
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\displaystyle x\left ( \log x \right )^{2}+2x \log x+x
Explanation
\displaystyle I= \int \left ( \log x \right )^{2}dx= \int 1.\left ( \log x \right )^{2}dx
\displaystyle = \left ( \log x \right )^{2}.x- \int 2\left ( \log x \right ).\frac{1}{x}.x\:dx
\displaystyle = x\left ( \log x \right )^{2}-2 \int 1.\log x dx
\displaystyle = x\left ( \log x \right )^{2}-2 \left [ x\log x-\int \frac{1}{x}.xdx \right ]
\displaystyle = x\left ( \log x \right )^{2}-2x \log x+2x
Hence, option 'B' is correct.
\displaystyle \int \left [ \left ( 1+x \right )e^{x}f\left ( x \right )+xe^{x}f'\left ( x \right ) \right ]dx=e^{x},
then
\displaystyle f\left ( x \right )=
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1
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x
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1/x
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\displaystyle e^{x}
Explanation
\displaystyle \frac{d}{dx}\left ( xe^{x} \right )=\left ( 1+x \right )e^{x}
\displaystyle \therefore I=\int f\left ( x\right )dx+ \int xe^{x}f'\left ( x \right )dx
Integrating 1st by parts
\displaystyle xe^{x}f\left ( x \right )-\int xe^{x}f'\left ( x \right )x+\int xe^{x}f'\left ( x \right )dx= xe^{x}f\left ( x \right )=e^{x}\therefore xf\left ( x \right )=1
or
\displaystyle f\left ( x \right )=1/x\Rightarrow \left ( c \right ).
\displaystyle \int \cos \sqrt{x}dx.
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\displaystyle \left [ \sqrt{x}\sin \sqrt{x}+\cos \sqrt{x} \right ].
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\displaystyle 2\left [ \sin \sqrt{x}-\cos \sqrt{x} \right ].
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\displaystyle 2\left [ \sqrt{x}\sin \sqrt{x}+\sqrt{x}\cos \sqrt{x} \right ].
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\displaystyle 2\left [ \sqrt{x}\sin \sqrt{x}+\cos \sqrt{x} \right ].
Explanation
\displaystyle I=\int \cos\sqrt{x}dx.
Put
\displaystyle \sqrt{x}=t
so that
\displaystyle \frac{1}{2\sqrt{\left ( x \right )}}dx=dt
\displaystyle \therefore I=f \left ( \cos t \right )2t\:dt=2\int t\cos t\:dt
Integrating by parts, we get
\displaystyle I=2\left [ t\sin t-\int 1.\sin t\:dt \right ]
\displaystyle =2\left [ t\sin t+\cos t \right ]
\displaystyle =2\left [ \sqrt{x}\sin \sqrt{x}+\cos \sqrt{x} \right ].
Hence, option 'D' is correct.
\displaystyle I=\int e^{x}\left [ \frac{1+\sqrt{1-x^{2}}\sin ^{-1}x}{\sqrt{\left ( 1-x^{2} \right )}} \right ]dx
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-e^{-x}\sin ^{-1}x+c
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-e^{x}\cos ^{-1}x+c
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e^{-x}\cos ^{-1}x+c
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e^{x}\sin ^{-1}x+c
Explanation
\displaystyle I=\int e^{x}\left [ \frac{1+\sqrt{1-x^{2}}\sin ^{-1}x}{\sqrt{\left ( 1-x^{2} \right )}} \right ]dx
\displaystyle =\int e^{x}.\frac{1}{\sqrt{\left ( 1-x^{2} \right )}}dx+\int e^{x}\sin^{-1}xdx.
Integrate 1st by parts
\displaystyle I=e^{x}.\sin^{-1}x-\int e^{x}\sin ^{-1}xdx+\int e^{x}\sin^{-1}xdx\:I=e^{x}\sin ^{-1}x.
The last two integrals cancel.
Note : Above is also of the form
\displaystyle \int e^{x}\left [ f\left ( x \right )+f'\left ( x \right )dx=e^{x}f\left ( x \right ) \right ]
\displaystyle f\left ( x \right )=\sin^{-1}x
and
\displaystyle f'(x)=\frac{1}{\sqrt{\left ( 1-x^{2} \right )}}.
\displaystyle\int \frac{x}{1+\cos x}dx.
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\displaystyle x\tan \left ( x/2 \right )-2\log \sec \left ( x/2 \right ).
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\displaystyle \tan \left ( x/2 \right )-2\log \sec \left ( x/2 \right ).
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\displaystyle x\tan \left ( x/2 \right )+2\log \sec \left ( x/2 \right ).
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\displaystyle x\tan \left ( x/2 \right )-\log \sec \left ( x/2 \right ).
Explanation
\displaystyle I=\int \frac{x}{1+\cos x}dx=\int \frac{x}{2\cos ^{2}\left ( x/2 \right )}dx
\displaystyle \int x\left [ \frac{1}{2}\sec^{2}\frac{x}{2} \right ]dx=x\tan\frac{x}{2}-\int 1.\tan \frac{x}{2}dx
\displaystyle =x\tan \left ( x/2 \right )-2\log \sec \left ( x/2 \right ).
Hence, option 'A' is correct.
\displaystyle\int \sin x\log \left ( \sec x+\tan x \right )dx.
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\displaystyle \cos x \log \left ( \sec x+\tan x \right )-x^2+c
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\displaystyle \sin x \log \left ( \sec x+\tan x \right )-x+c
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\displaystyle -\cos x \log \left ( \sec x+\tan x \right )-x+c
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\displaystyle -\sec x \log \left ( \sec x+\tan x \right )-x^2+c
Explanation
\displaystyle I= \int \sin x\log \left ( \sec x+\tan x \right )dx
\displaystyle I= \left ( -\cos x \right )\log \left ( \sec x+\tan x \right )-\int \left [ \frac{d}{dx}\log\left ( \sec x+\tan x \right ). \left ( -\cos x \right ) \right ]dx.
Now
\displaystyle \int \sec xdx=\log \left ( \sec x+\tan x\right );
\displaystyle \therefore \frac{d}{dx}\log \left (\sec x+\tan x \right )=\sec x
\displaystyle \therefore I=-\cos x \log \left ( \sec x+\tan x \right )-\int \sec x\left ( -\cos x \right )dx
\displaystyle =-\cos x \log \left ( \sec x+\tan x \right )-\int 1dx
\displaystyle =-\cos x \log \left ( \sec x+\tan x \right )-x +c
Hence, option 'C' is correct.
Solve
\displaystyle\int e^{x}\frac{1-\sin x}{1-\cos x}dx
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\displaystyle -e^{x}\cot \left (\dfrac {x}{2} \right )
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\displaystyle -e^{x}\tan \left (\dfrac{x} {2} \right )
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\displaystyle e^{x}\tan \left (\dfrac{x} {2} \right )
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\displaystyle e^{x}\cot \left (\dfrac {x}{2} \right )
Explanation
\displaystyle I=\int e^{x}\frac{\left ( 1-\sin x \right )}{\left ( 1-\cos x \right )}dx
\displaystyle I=\int \left [ \frac{e^{x}}{2\sin^{2}\left ( x/2 \right )}-e^{x}\frac{2\sin \left ( x/2 \right )\cos \left ( x/2 \right )}{2/sin^{2}\left ( x/2 \right )} \right ]dx
\displaystyle =\int e^{x}\left ( \frac{1}{2}co\sec^{2}\frac{x}{2} \right )dx-\int e^{x}\cot \frac{x}{2}dx.
Integrate first by parts
\displaystyle \therefore I=e^{x}\left ( -\cot \frac{x}{2} \right )-\int e^{x}\left ( -\cot \frac{x}{2} \right )dx-\int e^{x}\cot \frac{x}{2}dx
\displaystyle= -e^{x}\cot \left (\dfrac {x}{2} \right )
The last two integrals cancel.
\displaystyle \int 4^{x}\left [ g'\left ( x \right )+g\left ( x \right )\log 4\right ]dx=
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\displaystyle \frac{4^{x}}{\log 4}g\left ( x \right )
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\displaystyle 4^{x}
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\displaystyle 4^{x}\log 4.g\left ( x \right )
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\displaystyle 4^{x}g\left ( x \right )
Explanation
\displaystyle \int 4^{x}\left [ g'\left ( x \right )+g\left ( x \right )\log 4\right ]dx
\displaystyle I=\int 4^{x} g'\left ( x \right )dx+\int g\left ( x \right)4^{x}\log 4dx
=4^{ x }g(x)-\int { (4^{ x } } \log 4\int { g'\left( x \right) } dx)dx+\int g\left( x \right) 4^{ x }\log 4dx
=4^{ x }g(x)-\int { (4^{ x } } \log 4g(x)dx+\int g\left( x \right) 4^{ x }\log 4dx
\displaystyle \therefore I=4^{x}g\left ( x \right )+C
\displaystyle \int \frac{x}{1+\sec x}dx.
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\displaystyle \frac{x^{2}}{2}+x\tan \frac{x}{2}+2\log \sec \frac{x}{2}
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\displaystyle \frac{x^{2}}{2}-x\tan \frac{x}{2}+2\log \sec \frac{x}{2}
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\displaystyle \frac{x^{2}}{2}-\tan \frac{x}{2}+2\log \sec \frac{x}{2}
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\displaystyle \frac{x^{2}}{2}-x\tan \frac{x}{2}-2\log \sec \frac{x}{2}
Explanation
\displaystyle I=\int \frac{x}{\sec x+1}dx= \int \frac{x\cos x}{1+\cos x}dx
\displaystyle =\int x\frac{1+\cos x-1}{1+\cos x}dx= \int \left ( x-\frac{x}{2\cos^{2}\left ( x/2 \right )} \right )dx
\displaystyle I=\frac{x^{2}}{2}-x\tan \frac{x}{2}+2\log \sec \frac{x}{2}
\displaystyle \int \frac{co\sec^{2}x-203}{\left ( \cos x \right )^{203}}dx
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\displaystyle \frac{-\cot x}{\left ( \cos x \right )^{203}}
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\displaystyle\frac{\cos x}{\left ( co\sec x \right )^{203}}
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\displaystyle\frac{-\tan x}{\left ( co\sec x \right )^{203}}
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\displaystyle\frac{\cot x}{\left ( \sin x \right )^{203}}
Explanation
\displaystyle \int \left ( \cos^{-203} x\right )co\sec^{2}xdx-\int \frac{203}{\left ( \cos x \right )^{203}}dx
\displaystyle =I_{1}-I_{2}
Integrate
\displaystyle I_{1}
by parts
\displaystyle \therefore I_{1}=\left ( \cos ^{-203}x \right )\left ( -\cot x \right )-\int \cot x.203\cos ^{-204}x\left ( -\sin x \right )dx
\displaystyle I_{1}=-\frac{\cot x}{\left ( \cos x \right )^{203}}+\int \frac{203}{\left ( \cos x \right )^{203}}dx +C
\displaystyle \therefore I_{1}-I_{2}=-\frac{\cot x}{\left ( \cos x \right )^{203}}+C
\displaystyle \int a^{x}\left [ \log x+\log a\log \left ( \frac{x^{x}}{e^{x}} \right ) \right ]dx
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\displaystyle x\left ( \log x-1 \right )
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\displaystyle a^{x}.x\left ( \log x-1 \right )
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\displaystyle a^{x}.x\left ( \log x+1 \right )
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\displaystyle a^{x}.\left ( \log x-1 \right )
Explanation
\displaystyle I= \int a^{x}\log x dx+\int \left ( a^{x}\log a \right ).x\left ( \log x-1 \right )dx
We know that
\displaystyle \int \log x dx= x\log x -\int x.\frac{1}{x}dx=x\left ( \log x-1 \right )
Integrating 1st term by parts,
\displaystyle \therefore I=a^{x}.x\left ( \log x-1 \right )-\int x\left ( \log x-1 \right )\left ( a^{x}\log a \right )dx+\int \left ( a^{x}.\log a \right ).x\left ( \log x-1 \right )dx
\displaystyle =a^{x}.x\left ( \log x-1 \right )
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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