Integration by Substitution
Trending Questions
Q.
What is the integral of ?
Q.
None of these.
Q.
If , find the value of .
Q. Let S(x)=∫dxex+8e−x+4e−3x, R(x)=∫dxe3x+8ex+4e−x and M(x)=S(x)−2R(x). If M(x)=12tan−1(f(x))+c then f(0)=
- 32
- 12
- 52
- 72
Q. ∫x.(xx)x.(2 log x+1)dx is equal to
- x(xx)+C
- (xx)x+C
- xx.log x+C
- None of these
Q. If ∫(3sin x+4)(3+4sin x)2=f(x)+C, wheref(0)=−13 and f(π2)=0( C is constant of integration), then
- The minimum value of |f(x)| is 13
- The minimum value of |f(x)| is 0
- The value of f(π) is 13
- The value of f(π) is 0
Q. Evaluate ∫sin−1(2x+2√4x2+8x+13) dx
- (x+1)tan−1(2x+23)−34log(4x2+8x+13)+c
- (x+1)tan−1(23)−34log(x2+8x+13)+c
- tan−1(2x+23)+34log(4x2+8x+13)+c
- none of these
Q. The value of the definite integral 1∫−1dx(1+ex)(1+x2) is
- π4
- π8
- π2
- π16
Q. ∫2x+3√3−xdx is equal to (where C is integration constant)
- 18√3−x+23(3−x)32+C
- −18√3−x−43(3−x)32+C
- −18√3−x+43(3−x)32+C
- −18√3−x−23(3−x)32+C
Q.
Let I =∫exe4x+e2x+1dx.J=∫e−xe−4x+e−2x+1dx, Then, for an arbitrary constant c, the value of J-I equals
12log(e4x−e2x+1e4x+e2x+1)+c
12log(e2x+ex+1e2x−ex+1)+c
12log(e2x−e2x+1e2x+ex+1)+c
12log(e4x−e2x+1e4x−e2x+1)+c
Q. The integral ∫e3loge2x+5e2loge2xe4logex+5e3logex−7e2logex dx, x>0, is equal to :
(where c is a constant of integration)
(where c is a constant of integration)
- loge|x2+5x−7|+c
- 14loge|x2+5x−7|+c
- 4loge|x2+5x−7|+c
- loge√x2+5x−7+c
Q. The value of ∫dxxn(1+xn)1n, nϵN
- 11−n{1+1xn}1−1n+c
- 11+n{1−1xn}1−1n+c
- −11−n{1−1xn}1−1n+c
- 11−n{1+1xn}1−1n+c
Q. If ∫x4+1x6+1 dx=tan−1(f(x))−23tan−1(g(x))+c, where c is arbitrary constant, then which of the following is/are correct?
- f(x) is odd function.
- g(x) is odd function.
- f(x)=g(x) has no real roots
- ∫f(x)g(x) dx=−1x+13x3+c1
(c1 is a arbitrary constant)
Q. ∫cos3x+cos5xsin2x+sin4xdx equals
- sinx−6tan−1(sinx)+c
- sinx−2sin−1x+c
- sinx−2(sinx)−1−6tan−1(sinx)+c
- sinx−2(sinx)−1+5tan−1(sinx)+c
Q. If ∫xcosα+1(x2+2xcosα+1)3/2dx=f(x)√g(x)+1+C, then g(x)−2cosαf(x) is
(where C is constant of integration)
(where C is constant of integration)
- x2
- x
- 2x2
- 2x
Q. ∫sin−1(2x+2√4x2+8x+13) dx=f(x)tan−1(g(x))−34ln(h(x))+k
If h(−1)=9, then which of the following option(s) is/are correct?
( k is a constant of integration )
If h(−1)=9, then which of the following option(s) is/are correct?
( k is a constant of integration )
- f(x)=23(x+1)
- g(x)=2x+23
- Minimum value of h(x) is 9
- ddxh(x)=8(x+1)
Q. ∫dxcosx+√3sinx equals
- logtan(x2+π12)+c
- logtan(x2−π12)+c
- 12logtan(x2+π12)+c
- 12logtan(x2−π12)+c
Q. If the integral of the function sin(lnx)x=f(x), then find the value of f(1)
(take constant of integration equal to zero)
___
(take constant of integration equal to zero)
Q. If ∫x+1√2x−1dx=f(x)√2x−1+C, where C is a constant of integration, then f(x) is equal to :
- 13(x+1)
- 23(x+2)
- 23(x−4)
- 13(x+4)
Q. If f(x)=∫5x8+7x6(x2+1+2x7)2dx, (x≥0), and f(0)=0, then the value of f(1) is :
- −12
- 14
- 12
- −14
Q. If ∫8x43+13.x38(x13+x5+1)4dx=13.xa(xb+xc+1)3+C where a, b, c ϵ N, (a>b>c and where C is a constant of integration), then
- a+b=39
- a+b=52
- a+b+c=57
- b−c=8
Q. ∫dxx12(1+x2)5/4 is equal to
- −2√x4√√1+x2+C
- 2√x4√√1+x2+C
- −√x4√√1+x2+C
- √x4√√1+x2+C
Q. ∫f(x)dx=ψ(x), then ∫x5f(x3)dx is equal to
- 13[x3ψ(x3)−∫x2ψ(x3)dx]+c
- 13x3ψ(x3)−∫x3ψ(x3)dx+c
- 13x3ψ(x3)−∫x2ψ(x3)dx+c
- 13[x3ψ(x3)−∫x3ψ(x3)dx]+c
Q. The integral ∫2x3−1x4+x dx is equal to : (Here C is a constant of integration)
- loge∣∣∣x3+1x∣∣∣+C
- 12loge(x3+1)2|x3|+C
- loge∣∣x3+1∣∣x2+C
- 12loge∣∣x3+1∣∣x2+C
Q. The integral ∫sec2x(secx+tanx)9/2dx equals
- −(secx−tanx)11/2{111+17(secx−tanx)2}+K
- 1(secx+tanx)11/2{111+17(secx+tanx)2}+K
- −1(secx+tanx)11/2{111+17(secx+tanx)2}+K
- (secx−tanx)11/2{111+17(secx−tanx)2}+K
Q. ∫e√x√x(x+√x)dx equals
- 2e√x[x−√x+1]+C
- e√x[x−2√x+1]
- e√x[x+√x]+C
- e√x[x+√x+1]+C
Q. The integral e∫1{(xe)2x−(ex)x}logex dx is equal to:
- 32−1e−12e2
- 32−e−12e2
- −12+1e−12e2
- 12−e−1e2
Q. ∫sin(tan−1√x) dx (x≥0) is
- √x2+x−12ln{(x+12)+√x2+x}+c
- 2√x2+x−12ln{(x+12)+√x2+x}+c
- √x2+x−ln{(x+12)+√x2+x}+c
- √x2+x+ln{(x+12)+√x2+x}+c
Q. The value of the integral ∫cos3x+cos5xsin2x+sin4x dx is
- sin x−6tan−1(sin x)+c
- sin x−2(sin x)−1+c
- sin x−2(sin x)−1−6tan−1(sin x)+c
- sin x−2(sin x)−1+5tan−1(sin x)+c
Q. The integral ∫2x3−1x4+x dx is equal to : (Here C is a constant of integration)
- loge∣∣∣x3+1x∣∣∣+C
- 12loge(x3+1)2|x3|+C
- loge∣∣x3+1∣∣x2+C
- 12loge∣∣x3+1∣∣x2+C