Integration by Parts
Trending Questions
Find the value of ∫∞0x e−xdx
1
-1
0
None of these
ILATE rule should always be applied whenever we are using integration by parts
True
False
Find the value of tan 3A - tan2A - tanA is equal to ________
tan 3A tan2A tanA
tan A tan2A - tan2A + tan3A - tan3AtanA
tan 3A + tan2A
-tan 3A tan 2A tanA
Let A{2, 3, 5, 7}. Examine whether the statements given below are true or false.
(i) ∃ x∈A such that x+3>9.
(ii) ∃ x∈A such that x is even.
(iii) ∃ x∈A such that x+2=6.
(iv) ∀ x∈A, x is prime.
(v) ∀ x∈A, x+2<10
(vi) ∀ x∈A, x+4≥11
- 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
∫[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)
- 12(√x2−1x3)+tan−1√x2−1+c
- 12(√x2−1x2)+xtan−1√x2−1+c
- 12(√x2−1x)+tan−1√x2−1+c
- 12(√x2−1x2)+tan−1√x2−1+c
- x sin−1(x)+√1−x2+C
- −(x sin−1(x)+√1−x2)
- x sin−1 x+√1−x2+C
- None of these
- (x−1)ex+1x+c
- xex+1x+c
- (x+1)ex+1x+c
- −xex+1x+c
- 3π2
- π2
- π3
- 2π3
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
- cos180
- 2cos180
- sin180
- 2sin180
- x[cos(logex)+sin(logex)]+C
- x[cos(logex)−sin(logex)]+C
- x2[sin(logex)−sin(logex)]+C
- x2[cos(logex)+sin(logex)]+C
- (3e+2)/6
- (3e−2)/6
- { (3e+2)2/36 }
- None of these.
- 3ln(2−sinϕ)4(sinϕ−2)+c
- 3ln(sinϕ−2)+42−sinϕ+c
- ln(2−sinϕ)+42−sinϕ+c
- 3ln(2−sinϕ)+4(2−sinϕ)+c
- f(x)=sinx
- f(x)=cosx
- A=π4
- A=π2
- e−3x(4cos4x−3sin4x)25+c
- −e−3x(3cos4x+4sin4x)25+c
- −e−3x(4cos4x+3sin4x)25+c
- None of these
(A) 1 (B) 2 (C) 3 (D) None.
- √a2+b2
- (a2+b2)13
- a2+b2
- (a2−b2)
- cosx logx+C
- sinx logx+C
- cosx log x +C