Integration by Parts
Trending Questions
- xtan−1(x)−xln|1+x2|+c
- x2tan−1(x)−12ln|1+x2|+c
- xtan−1(x)−12ln|1+x2|+c
- xtan−1(x)−x2ln|1+x2|+c
Why Is Integration Antiderivative?
- sin 2x In (tan x) – 2x + C
- sin 2xIn (tan x) + 2x + C
- sin xIn (tan x) – x + C
- none of these
- (x−1)ex+1x+c
- xex+1x+c
- (x+1)ex+1x+c
- −xex+1x+c
- sin x
- cos x
- tan x
- cot x
- esinx(tanx+x)+C
- esinx(x−secx)+C
- esinx(secx+tanx)+C
- None of these
- 12esin2x(3−sin2x)
- 12esin2x(1−12cos2x)
- esin2x(3cos2x+2sin2x)
- esin2x(2cos2x+3sin2x)
- In has finite value for all natural values of 'n'
- I2I1+I3I2+I4I3=−4
- I4=−9
- I8=40331
(where ′C′ is the constant of integration)
- sin(59x)sin58x58+C
- sin(58x)sin59x60+C
- sin(60x)sin59x59+C
- sin(59x)sin59x59+C
- cosx.e(sinx−1x)+c
- 1x.e(sin−1x)+c
- x.e(sin x−1x)+c
- e(sinx−1x)+x+c
- sin 3x
- sin 2x
- sin x
- sin 4x
and ∫ln20e−2xf(x)dx=3, then∫ln20e−2xf′′(x)dx
is
- False
- True
Im, n=∫10xm(logx)ndx, then Im, n is equal to
nn+1Im, n−1
-mn+1Im, n−1
-nm+1Im, n−1
mn+1Im, n−1
- S∞=π2
- S3=π6
- S∞=π4
- S3=π3
Im, n=∫10xm(logx)ndx, then Im, n is equal to
nn+1Im, n−1
-mn+1Im, n−1
-nm+1Im, n−1
mn+1Im, n−1
- −2x3−1
- −4x3−1
- 4x3+1
- −2x3+1
∫[f(x)g′′(x)−f"(x)g(x)]dx is equal to
f(x)g′(x)
f′(x)g(x)−f(x)g′(x)
f(x)g′(x)−f′(x)g(x)
f(x)g′(x)+f′(x)g(x)
If ∫g(x)dx=g(x), then ∫g(x){f(x)+f′(x)}dx is equal to
g(x)f(x)−g(x)f′(x)+C
g(x)f′(x)+C
g(x)f(x)+C
g(x)f2(x)+C
- xcosx−sinx
- xcosx+sinx
- −xcosx−sinx
- −xcosx+sinx
- 10tan−1(10)−12ln|102|
- 10tan−1(1+102)−12ln|102|
- tan−1(10)−12ln|1+102|
- 10tan−1(10)−12ln|1+102|
- (2, 2e2)
- (2, e2)
- (2, 3e2)
- (2, 3e2)
- 13[x3ψ(x3)−∫x2ψ(x3)dx]+c
- 13x3ψ(x3)−3∫x3ψ(x3)dx+c
- 13x3ψ(x3)−∫x2ψ(x3)dx+c
- 13[x3ψ(x3)−∫x3ψ(x3)dx]+c
In some of the cases we can split the integrand into the sum of the two functions such that the integration of one of them by parts produces an integral which cancels the other integral. Suppose we have an integral of the type
∫[f(x)h(x)+g(x)]dx
Let ∫f(x)h(x)dx=I1 and ∫g(x)dx=I2
Integrating I1 by parts, we get
I1=f(x)∫h(x)dx−∫{f′(x)∫h(x)dx}dx
∫xex(1+x)2dx is equal to
xex+c
ex(x+1)2+c
ex1x+1+c
exx+1+c
- λ=4
- μ=2
- k=1
- γ=2