Explanation
The value of ∫lnn(1−(1x))dxx(x−1) is
Consider, I=∫ln(1−1x)x(x−1)dx
I=∫ln(1−1x)x2(1−1x)dx
Put ln(1−1x)=t ⇒ 1(1−1x)×1x2dx=dt
I=∫tdt
I=t22+c
I=12[ln(1−1x)]2+c
Consider the given equation.
I=∫etan−1x(1+x+x21+x2)dx
I=∫(etan−1x+xetan−1x+x2etan−1x1+x2)dx ……… (1)
Let t=xetan−1x
t=xetan−1x
dtdx=etan−1x×1+x(etan−1x)×11+x2
dtdx=etan−1x+xetan−1x1+x2
dt=(etan−1x+x2etan−1x+xetan−1x1+x2)dx
From equation (1), we get
I=∫1dt
I=t+C
On putting the value of t, we get
I=xetan−1x+C
Hence, this is the answer.
Consider the given integral.
I=∫x√x2+2dx
Let t=x2+2
dtdx=2x+0
dt2=xdx
Therefore,
I=12∫dt√t
I=12(2√t)+C
I=√t+C
I=√x2+2+C
Consider the following integral.
∫2x√1−4xdx
Solution:
Let and differentiate w.r.t “x” we get.
t=2x
dtdx=2xln2
∫2x√1−4xdx......(1)
Put the value dx of in eq . (1) we get.
=∫2x√1−4xdx......(1)
=∫2x2xln2√1−(t2)xdx
=∫1ln2√1−(t2)xdx
=1ln2sin−1x+C
Hence, the value of [K=1ln2]
Hence, this is the correct answer .
I=∫cos(logx)dx
Let t=logx
dtdx=1x
xdt=dx
I=∫etcostdt
I=costet−∫(−sint)etdt
I=etcost+∫(sint)etdt
I=etcost+sintet−∫costetdt
I=etcost+sintet−I
2I=et(cost+sint)+C
I=et2(cost+sint)+C
On putting the value of ′t′, we get
I=elogx2(cos(logx)+sin(logx))+C
I=x2(cos(logx)+sin(logx))+C
Consider the following integral .
I=∫xsinxsec3xdx=∫xtanxsec2x
let t=tanx and differentiate both side w.r.t x, we get.
dtsec2x=dx
=∫ttan−1tsec2xsec2xdt
=∫ttan−1tdt
=tan−1t∫tdt−12∫11+t2t2dt
=tan−1t∫tdt−12(∫t2+1−11+t2dt)
=tan2x2tan−1(tanx)+tan2x2tan−1(tanx)−tanx2+C
=tan2xtan−1(tanx)−tanx2+C
Hence, this is the correct answer.
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