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CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 12 - MCQExams.com

If I=exlog(ex+1)dx, then I equals
  • x+(ex+1)log(ex+1)+C
  • x+(ex+1)log(ex+1)+C
  • x(ex+1)log(ex+1)+C
  • none of these
If I =5x2+xdx, then I equals
  • x+25x+3sin1x+23+C
  • x+25x+3sin1x+27+C
  • x+25x+5sin1x+25+C
  • None of these
cos(lnx)dx=
  • x2(coslnx+sinlnx)
  • x2[coslnxsinlnx]
  • [xcoslnx+sinlnx]
  • None of these
The evaluation of pXp+2q1qXq1X2p+2q+2Xp+q+1dx is 
  • XpXp+q+1+C
  • XqXp+q+1+C
  • XqXp+q+1+C
  • XpXp+q+1+C
xsinxsec3xdx is equal to
  • 12[sec2xtanx]+c
  • 12[xsec2xtanx]+c
  • 12[xsec2x+tanx]+c
  • 12[sec2x+tanx]+c
x2(xsec2x+tanx)(xtanx+1)2dx
  • x2[1xtanx+1]+2log(xsinx+cosx)+C
  • x2[1xtanx+1]+2log(sinx+xcosx)+C
  • x[1xtanx+1]+2log(sinx+xcosx)+C
  • x[1xtanx+1]+2log(xsinx+cosx)+C
Integrate xa2x2 with respect to x.
  • (a2x2)+C 
  • a2x2+C
  • (a2x2)323+C
  • None of the above
e(x2+4lnx)x3ex2x1dx is equal to
  • (e2lnx12)ex2+C
  • (x1)xex22+C
  • (x21)2xex2+C
  • none of these
sin2x.logcosxdx is equal to
  • cos2x(12+logcosx)+k
  • cos2x.logcosx+k
  • cos2x(12logcosx)+k
  • none of these
If f(x)dx=F(x), then x3f(x2)dx equals to
  • 12[x2(F(x))2(F(x))2dx]
  • 12[x2F(x2)F(x2)d(x2)]
  • 12[x2F(x)12(F(x))2dx]
  • 12[x2F(x2)+F(x2)d(x2)]
If xlog(1+x2)dx=ϕ(x).log(1+x2)+Ψ(x)+c then
  • ϕ(x)=1+x22
  • Ψ(x)=1+x22
  • Ψ(x)=1+x22
  • ϕ(x)=1+x22
(x2+1)[ln(x2+1)2lnxx4]dx;g(x)=(1+1x2); equals
  • 13(g(x))1/2ln(g(x))+29(g(x))+C
  • 13(g(x))3/2ln(g(x))+29(g(x))3/2+C
  • 13(g(x))3/2ln(g(x))29(g(x))+C
  • None of these
etan1x(1+x2)[(sec11+x2)2+cos1(1x21+x2)dx] at (x>0)
  • etan1x.tan1x+C
  • etan1x.(tan1x)22+C
  • etan1x.(sec1(1+x2))2+C
  • etan1x.(cosec1(1+x2))2+C
[log(1+cosx)xtan(x2)]dx
  • xlog(1+tanx)
  • xlog(1+sinx)
  • log(1+sinx)
  • xlog(1+cosx)
2x+sin2x1+cos2xdx 
  • xcotx.
  • xtanx.
  • x2tanx.
  • xsecx.
If I=10et1+t dt, then p=10etlog(1+t)dt=
  • I
  • 2I
  • elog2I
  • none
Eavaluate :  xtan1x  
  • 12(x2+1)tan1x12x+C
  • 12x2tan1x12x+C
  • 12x2tan1x+12x+C
  • none of these
If (x32x2+3x1)cos2xdx
=sin2x4u(x)+cos2x8v(x)+C then
  • u(x)=x34x2+3x
  • u(x)=2x34x2+3x
  • u(x)=3x34x+3
  • v(x)=6x28x
Repeated application of integration by parts gives us the reduction formula, if the integrand is dependent on a natural number n.

If dxxnax+b=ax+b(n1)bxn1Adxxn1ax+b+C, then A is equal to
  • 2n32n1
  • 2n32n2ab
  • n2n1ab
  • n3n2ab
If I=sec2xcosec4xdx=Kcot3x+Ltanx+Mcotx+C then

  • K=1/3
  • L=2
  • M=2
  • none of these
If xex1+exdx=f(x)1+ex2logg(x)+C, then
  • f(x)=x1
  • g(x)=1+ex11+ex+1
  • g(x)=1+ex+11+ex1
  • f(x)=2(x2)
If log(1x+1+x)dx=xf(x)+Lx+Msin1x+C, then
  • f(x)=log(1x+1+x)
  • L=12
  • M=23
  • M=12
If (x32x2+3x1)cos2xdx=sin2x4u(x)+cos2x8v(x)+c, then
  • u(x)=x34x2+3x
  • u(x)=2x34x2+3x
  • v(x)=3x24x+3
  • v(x)=k6x28x
[x2+1[ln(x2+1)2lnx]x4]
  • (x21)x219x3[23ln(1+1x2)]
  • (x21)x219x3[2+3ln(1+1x2)]
  • (x2+1)x2+19x3[23ln(1+1x2)]
  • (x21)x2+19x3[2+3ln(1+1x2)]
If I=sec1xcosec1xsec1x+cosec1xdx, then I equals
  • (4/π)(xsec1xx1)+x+C
  • (4/π)(xsec1x+x1)x+C
  • (4/π)(xsec1xx1)x+C
  • none of these
If xe5x2sin4x2dx=Ke5x2(Asin4x2+Bcos4x2)+C. Then
  • 182,5,4
  • 182,5,4
  • 182,5,4
  • 182,15,14
ex4(x+x3+2x5)ex2dx is equal to
  • 12xex2ex4+c
  • 12x2ex4+c
  • 12ex2ex4+c
  • 12x2ex2ex4+c
If I=(1x1/2x1/4+log(1+x1/4)x1/2+x1/4)dx, then I equals
  • 2x+4x1/4+log|x1/41|+C
  • 2(x1/26x1/4+3)log(x1/4+1)+C(x1/210x1/4+7)+2(log(x1/4+1))2
  • xx1/4+6log(x1/4+1)3x3/4+C
  • none of these
sin2(lnx)dx is equal to
  • 2x15(5+2sin(2lnx)+cos(2lnx))+c
  • x5(5+2sin(2lnx)cos(2lnx))+c
  • x10(52sin(2lnx)cos(2lnx))+c
  • x10(52sin(2lnx)+sin(2lnx))+c 
cos2θ.lncosθ+sinθcosθsinθdθ
  • 12ln(cosθsinθcosθ+sinθ)sin2θ12ln(sec2θ)+C
  • 12ln(cosθsinθcosθ+sinθ)sin2θ12ln(cosec2θ)+C
  • 12ln(cosθ+sinθcosθsinθ)sin2θ12ln(cosec2θ)+C
  • 12ln(cosθ+sinθcosθsinθ)sin2θ12ln(sec2θ)+C
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers