Integration by parts is a technique used to integrate the product of two functions.
Formula:
\int u\,dv = uv - \int v\,du
Solved Example
Question 1. Evaluate ∫2x cos(x)dxÂ
Solution:Â
Given that, ∫2x cos(x)dx
According to the rule of integration by parts,Â
Let us consider,
u = 2x and dv/dx = cos(x)Â
Then
du/dx = 2  and  v = ∫cos(x) =  sin(x)
Now, using the formula for integration by parts;
∫u(dv/dx)dx = uv - ∫v(du/dx)dx
We get,
∫2x cox(x)dx = 2x sin(x) - ∫sinx.2dx
∫2x cos(x)dx = 2x sin(x) + 2cos(x) + c
where c is a constant.
Question 2. Evaluate ∫3x ln|x| dx
Solution:
Given, ∫3xln|x| dxÂ
According to rule of integration by parts,Â
Let us consider,
u = ln|x| Â and dv/dx = 3x
Then
du/dx = 1/x  and  v = ∫3xdx = 3x2/2
Now, using the formula for integration by parts,
∫u(dv/dx)dx = uv - ∫v(du/dx)dx
∫3x ln|x|dx = ln|x|.3x2/2 - ∫3x2/2(1/x)dx
= 3x2/2(ln|x|) - ∫3x/2 dx
= 3x2/2(ln|x|) - 3x2/4 + C
where C is a constant.
Question 3. Evaluate ∫10x2
Solution:Â
Given that, ∫10x2
According to rule of integration by parts,Â
Let us consider,
u = 10x2 and dv/dx = 1
Then
du/dx = 20x  and  v = ∫(1)dx = xÂ
Now, using the formula for integration by parts;
∫u(dv/dx)dx = uv - ∫v(du/dx)dx
∫10x2 dx = 10x2x - ∫x(20x)dx
∫10x2 dx = 10x3 - ∫20x2dx
∫10x2 dx = 10x3 - 20/3x3 + C
∫10x2 dx = 10x3/3 + C
where C is a constant.
Question 4. Evaluate ∫x2 e4x dx
Solution:
Given that, ∫x2 e4x dx
According to rule of integration by parts,Â
Let us consider,
u = x2 and dv/dx = e4x
Then
du/dx = 2x  and  v = ∫e4x dx = 1/4(e4x)
Then, using the formula for integration by parts,
∫u(dv/dx)dx = uv - ∫v(du/dx)dx
∫x2e4x dx = (x2)1/4(e4x) - ∫1/4(e4x)(2x)dx
= 1/4(x2)(e4x) - ∫2/4(x)(e4x)dxÂ
= 1/4(x2)(e4x) - ∫1/2(x)(e4x)dxÂ
The resulting integral is still a product ofÂ
two functions i.e. 1/2(x) and e4x.Â
So, We use the formula for integration again.
This time we take
u = 1/2(x)  and  dv/dx = e4x
So
du/dx = 1/2  and  v = ∫e4xdx = 1/4(e4x)
Now, on putting these values in above equation,
∫x2e4x dx = 1/4(x2)(e4x) - ∫2/4(x)(e4x)dx
= 1/4(x2e4x) - {2/4(x).1/4(e4x) - ∫(1/4(e4x).2/4 dx}
= 1/4(x2e4x) - 1/8(xe3x) + 1/8(e4x) + C
where C is a constant.
Question 5. Evaluate ∫ 2ex sin(x) dx
Solution:
Given that, ∫ 2exsin(x) dx
According to rule of integration by parts,Â
Let us consider,
u = 2ex  and  dv/dx = sin(x)
Then
du/dx = 2ex  and v = ∫sin(x)dx = -cos(x)
Now, using the formula for integration by parts,
∫u(dv/dx)dx = uv - ∫v(du/dx)dx
∫ 2exsin(x) dx = (2ex)(-cos(x)) - ∫(-cos(x))(2ex)dx
= -(2ex)cos(x) + ∫ 2ex cos(x)dx
The resulting integral is yet a product of two
functions i.e. 2ex and cos(x).Â
So, we use the formula for integration again.
This time we take
u = 2ex and  dv/dx = cos(x)
So
du/dx = 2ex and  v = ∫cos(x)dx = sin(x)
Now, on putting these values in the above equation, we get
∫ 2exsin(x) dx = -cos(x)(2ex) + ∫ 2ex cos(x)dx
= -(2ex)cos(x) + {2exsin(x) - ∫sin(x)(2ex)dx}
= -2excos(x) + 2exsin(x) - ∫2exsin(x)dx
Now, notice that the integral ∫exsin(x)dx in the result
is same as the integral in the given question.
So, let us assume this as I.
Therefore,
I = ∫ 2exsin(x) dx
So,
I = 2exsin(x) - 2excos(x) - I
2I = 2exsin(x) - 2excos(x)
2I = 2(exsin(x) - excos(x))
I = (exsin(x) - excos(x))
So,
∫ 2exsin(x) dx = exsin(x) - excos(x) + C
where C is a constant.
Question 6. Evaluate ∫2tan-1x dxÂ
Solution:
Given, ∫2tan-1x dxÂ
According to the rule of integration by parts,Â
Let us consider,
u = tan-1x  and  dv/dx = 2
Then
du/dx = 1/(1 + x2)  and  v = ∫(2)dx = 2xÂ
Now, using the formula for integration by parts,
∫u(dv/dx)dx = uv - ∫v(du/dx)dx
∫2tan-1x dx = tan-1x(2x) - ∫2x(1/(1+x2))dxÂ
let, p = (1 + x2)
So, dp/dx = 2xÂ
⇒ x = 1/2
Now, substituting the above values, we get
∫tan-1x dx = tan-1x(2x) - ∫(1/p)dx
= 2xtan-1x - ∫(1/p)dx
= 2xtan-1x - lnp
= 2xtan-1x - ln(1 + x2)Â
= 2xtan-1x - ln√(1 + x2) + C
where C is a constant.
Practice Question
Question 1: Evaluate: ∫x sinx dx
Question 2: Evaluate: ∫x2 sinx dx
Question 3: Evaluate: ∫x3exdx
Question 4: Evaluate: ∫x3cosx dx
Question 5. Evaluate: ∫xexcosx dx