In the previous section, we saw the method of Integration by parts. We saw some solved examples also. In this section, we will see a special case of this method.
Integral of the type $\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$
This can be calculated in 3 steps:
1. We have: $\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$
$\small{~=~\int{\left[e^x\,f(x)\right]dx}~+~\int{\left[e^x\,f'(x)\right]dx}}$
• There are two terms. We will denote the first one as I1.
2. Let us calculate I1:
(i) Assigning first and second functions:
♦ Let first function be: f(x) = $\small{f(x)}$
♦ Let second function be: g(x) = $\small{e^x}$
(ii) Finding A:
$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[e^x \right]dx}~=~e^x}$
(iii) $\small{\big[f(x) \left(A \right) \big]~=~\big[f(x) \, \left(e^x \right) \big]}$
• This is the first term.
(iv) $\small{f'(x)~=~f'(x)}$
(v) $\small{\int{\big[f'(x)\,\left(A \right) \big]dx}~=~\int{\big[f'(x)\,\left(e^x \right) \big]dx}}$
• This is second term.
(vi) So we get:
I1 = $\small{\text{First term - Second term}}$
$\small{~=~\big[f(x) \, \left(e^x \right) \big]~-~\bigg[\int{\big[f'(x)\,\left(e^x \right) \big]dx} \bigg]}$
3. So from (1), we get:
$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$
$\small{=~\int{\left[e^x\,f(x)\right]dx}~+~\int{\left[e^x\,f'(x)\right]dx}}$
= I1 + $\small{\int{\left[e^x\,f'(x)\right]dx}}$
= $\big[f(x) \, \left(e^x \right) \big]~-~\bigg[\int{\big[f'(x)\,\left(e^x \right) \big]dx} \bigg]$ + $\small{\int{\left[e^x\,f'(x)\right]dx}}$
= $\small{f(x) \, \left(e^x \right)~+~\rm{C}}$
Let us see some solved examples:
Solved Example 23.57
Find $\small{\int{\big[e^x \left[\sin x + \cos x \right]\big]dx}}$
Solution:
1. Let f(x) = $\small{\sin x}$
• Then $\small{f'(x)~=~\cos x}$
2. So the given integral is of the form:
$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$
3. Then the result is:
$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$
$\small{~=~e^x \, \sin x~+~\rm{C}}$
Solved Example 23.58
Find $\small{\int{\left[e^x\,\sec x (1 + \tan x) \right]dx}}$
Solution:
1. The given integral can be written as:
$\small{\int{\big[e^x \left[\sec x + \sec x \tan x \right]\big]dx}}$
Let f(x) = $\small{\sec x}$
• Then $\small{f'(x)~=~\sec x \tan x}$
2. So the given integral is of the form:
$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$
3. Then the result is:
$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$
$\small{~=~e^x \, \sec x~+~\rm{C}}$
Solved Example 23.59
Find $\small{\int{\left[\frac{x e^x}{(1+x)^2} \right]dx}}$
Solution:
1. The $\small{\left[\frac{x}{(1+x)^2} \right]}$ portion can be rearranged as shown below:
$\small{\frac{x}{(1+x)^2} = \frac{1+x-1}{(1+x)^2} = \big[\frac{1+x}{(1+x)^2}~+~\frac{(-1)}{(1+x)^2}\big] = \big[\frac{1}{1+x}~+~\frac{(-1)}{(1+x)^2}\big]}$
2. Now, $\small{\frac{(-1)}{(1+x)^2}}$ is the derivative of $\small{\frac{1}{1+x}}$.
(The reader may verify this by doing the differentiation)
3. So the given integral is of the form:
$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$
Where $\small{f(x) = \frac{1}{1+x}}$
3. Then the result is:
$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$
$\small{~=~e^x \, \left[\frac{1}{1+x} \right]~+~\rm{C}}$
Solved example 23.60
Find $\small{(i)~\int{\bigg[e^x \left[\tan^{-1}x \,+\,\frac{1}{1+x^2} \right]\bigg]dx}~~~~~~(ii)~\int{\left[\frac{(x^2 + 1)e^x}{(x+1)^2} \right]dx}}$
Solution:
Part (i):
1. Let f(x) = $\small{\tan^{-1} x}$
• Then $\small{f'(x)~=~\frac{1}{1 + x^2}}$
2. So the given integral is of the form:
$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$
3. Then the result is:
$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$
$\small{~=~e^x \, \tan^{-1} x~+~\rm{C}}$
Part (ii):
1. The $\small{\left[\frac{(x^2 + 1)}{(x+1)^2} \right]}$ portion can be rearranged as shown below:
$\small{\frac{x^2 + 1}{(x+1)^2} = \frac{x^2 + 1-1+1}{(x+1)^2} = \frac{x^2 -1+2}{(x+1)^2} = \frac{x^2 -1}{(x+1)^2} + \frac{2}{(x+1)^2}}$
$\small{= \frac{(x+1)(x-1)}{(x+1)^2} + \frac{2}{(x+1)^2}= \frac{x-1}{x+1} + \frac{2}{(x+1)^2}}$
2. Now, $\small{\frac{2}{(x+1)^2}}$ is the derivative of $\small{\frac{x-1}{x+1}}$.
(The reader may verify this by doing the differentiation)
3. So the given integral is of the form:
$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$
Where $\small{f(x) = \frac{x-1}{x+1}}$
3. Then the result is:
$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$
$\small{~=~e^x \, \left[\frac{x-1}{x+1} \right]~+~\rm{C}}$
The link below gives a few more solved examples:
Exercise 23.6
We
have completed a discussion on integration by parts.
In the next section, we will see some standard integrals.
Previous
Contents
Next
Copyright©2025 Higher secondary mathematics.blogspot.com
No comments:
Post a Comment