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CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 2 - MCQExams.com

sin1(cosx)dx is equal to
  • πx2+c
  • πx22+c
  • πxx22+c
  • πx+x22+c
What is dxx(1+lnx)n equal to (n1) ?
  • 1(n1)(1+lnx)n1+c
  • 1n(1+lnx)1n+c
  • n+1(1+lnx)n+1+c
  • 1(n1)(1+lnx)n1+c
Evaluate xsecx.tanxdx=
  • xsecx+log|tan(π/2+x/2)|+c
  • xsecxlog|tan(π/4+x/2)|+c
  • xsecxlog|tan(π/4+x)|+c
  • xsecx+log|tanx/2|+c
{(logx1)1+(logx)2}2dx is equal to
  • x(logx)2+1+c
  • xex1+x2+c
  • xx2+1+c
  • logx(logx)2+1+c
sec2x.cosec2xdx=
  • tanxcotx+c
  • tanx+cotx+c
  • tanx+cotx+c
  • secxtanx+c
xx1dx,x(0,π/2) equals
  • x(x1)+log(x+x1)+c
  • x(x1)log(x+x1)+c
  • x(x1)+log(xx1)+c
  • x(x+1)log(xx+1)+c
xe2x(1+x)dx equal to

  • xex2+c
  • (ex)22r
  • (1+x)22+c
  • (xex)22
The value of x223xdx=
  • x223x4log2x.23x+19.(log2)3+23x+127(log2)+c
  • x223x3log2x.23x+19.(log2)+23x+127(log2)+c
  • x223x3log223x+1.x32(log2)2+23x+133(log2)3+c
  • x223xlog223x+1.x32(log2)2+23x+1(log2)3+c
logxdx=
  • x(logxx)+c
  • x(logx)+c
  • x(logx1)+c
  • x(logx+x)+c
Evaluate : cot2x(cosec2x+cosecx)dx
  • xsinx+C
  • x+cosx+C
  • sinxx+C
  • 2tan(ax2)+c
[cosxxsinxlogx]dx=
  • (logx)sinx+c
  • (logx)(cosx)+c
  • 2(logx)(cosx)+c
  • (logx)sinx+c
Evaluate x2exdx=
  • ex(x22x+2)+c
  • ex(x2+2x+2)+c
  • x2+ex+c
  • ex(x2+x+2)+c
elogxcosxdx=
  • xsinxcosx+c
  • x2sinx+cos2x+c
  • xsinx+cosx+c
  • xsinx+sinx+c
xcosxlogxsinxx(logx)2dx is equal to
  • sinxlogx+c
  • logx+sinx+c
  • logxsinx+c
  • cosxlogx+c
The value of x(cosecx cotx)dx= is
  • xcosecxlog|tanx/2|+c
  • 2xcosecx+log|tanx2|+c
  • xcosecx2log|tanx/2|+c
  • xcotxlog|tanx2|+c
Solve  tan1xdx
  • xtanxlog|1+x2|+c
  • x tan1x12log|1+x2|+c
  • x  tan1x+log1+x2+c
  • xtanx+log|1+x2|+c
Evaluate log(x/e)(logx)2dx
  • logxx+c
  • xlogx+c
  • xlog(x)2+c
  • (logx)2x+c
x2cosh4xdx=
  • x24sinh4xx8cosh4x+sinh4x32+c
  • x24sinh4x+x8 cosh 4xsinh4x32+c
  • x24sinh4xx8 cosh 4xsinh4x32+c
  • x24sinh4x+x8 cosh 4xsinh4x64+
cot1(x1x+1)dx=
  • π4x+xcot1x12log(1+x2)+c
  • xcot1x+12log(1+x2)+c
  • π4x12log(1+x2)+c
  • π4x+xcot1x+12log(1+x2)+c
excosecx(1cotx)dx=
  • excotx+c
  • excosecxcotx+c
  • excosecx+c
  • excotx+c
x2axdx=
  • ax[x2loga2x(loga)2+2(loga)3]+c
  • ax[x2loga+2x(loga)22(loga)3]+c
  • ax[x2loga+2x(loga)2+2(loga)3]+c
  • ax(xloga1(loga)2)+c
lf polynomials P and Q satisfy ((3x1)cosx+(12x)sinx)dv=Pcosx+Qsinx+R (ignoring the constant of integration) then
  • P=3x2
  • Q=2+x
  • P=3(x1)
  • Q=3(x1)
Evaluate arcsinx1xdx =
  • 2[x1x arc sinx]+c
  • 2[x+1x arc sinx]+c
  • 2[x+1x arc cosx]+c
  • 2[x1x arc cosx]+c
ex(x25x+8) dx =exf(x)+c then f(x)
  • x25x+12
  • x2+7x+15
  • x27x15
  • x27x+15

Evaluate x(sec2x1)(sec2x+1)dx
  • xtanxlog|secx|+x22+c
  • xtanxlog|secx|x22+c
  • xtanxlog|secx2|+c
  • xtanxlog|secx2|+x22+c
ex23log2xdx=exf(x)+c, then f(x)=
  • x33x2+6x6
  • x33x26x3
  • x33x2+6x+6
  • x3+3x2+6x+6
x+sinx1+cosxdx=
  • xtanx2+c
  • xcotx2+c
  • xsinx2+c
  • xcosx2+c
cos1(1x)dx equal to
  • xsec1x+cosh1x+c
  • xsec1xcosh1x+c
  • xsec1xsin1x+c
  • xsec1x+sin1x+c
Evaluate: log10xdx
  • (x1)logex+c
  • loge10.xloge(xe)+c
  • log10e.xloge(xe)+c
  • 1x+c
Evaluate xex(x+1)2dx
  • ex[1x+1]+c
  • ex[1x+1]+c
  • exx+c
  • ex[1(x+1)2]+c
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