Question

How do you evaluate the integral $\int {\dfrac{{{x^4}}}{{1 + {x^2}}}dx}$ ?

Answer 1

We use the concepts of integrations to solve this problem. We also use partial fractions. Generally, the reverse process of differentiation is known as integration. And we also use some formulas like $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$ , to solve this problem.

Complete step by step solution:
First of all, we will find the partial fractions of the given term for which we need to find the integral.
So, to find it, let us first add and subtract 1 in the denominator.
$\Rightarrow \dfrac{{{x^4} - 1 + 1}}{{1 + {x^2}}}$
So, we can group the terms as $\dfrac{{({x^4} - 1) + 1}}{{1 + {x^2}}}$
So, we get,
$\Rightarrow \dfrac{{({x^4} - 1) + 1}}{{1 + {x^2}}} = \dfrac{{{x^4} - 1}}{{1 + {x^2}}} + \dfrac{1}{{1 + {x^2}}}$
And ${x^4} - 1 = {({x^2})^2} - {1^2}$
Which is of the form ${(a)^2} - {(b)^2}$ which is equal to $(a + b)(a - b)$
So, ${x^4} - 1 = ({x^2} + 1)({x^2} - 1)$
So, we get, $\dfrac{{({x^2} + 1)({x^2} - 1)}}{{1 + {x^2}}} + \dfrac{1}{{1 + {x^2}}}$
$\Rightarrow ({x^2} - 1) + \dfrac{1}{{1 + {x^2}}}$
We founded the partial fractions and now we need to integrate this.
$\Rightarrow \int {\left( {({x^2} - 1) + \dfrac{1}{{1 + {x^2}}}} \right)} dx$
$\Rightarrow \int {({x^2} - 1)dx + \int {\dfrac{1}{{1 + {x^2}}}dx} }$
Now let us integrate the terms separately.
And we know that, $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$ and $\int {({\text{constant)}}} dx = ({\text{constant)}}x$
These are standard results, which you need to remember for your future applications of integrals.
So, we get the integral as
$\Rightarrow \int {({x^2} - 1)dx = \dfrac{{{x^3}}}{3} - x} + c$ ------(1)
Now let us integrate the second term.
To integrate it, take
$x = \tan \theta$
$\Rightarrow dx = ({\sec ^2}\theta )d\theta$ $\left( {\because \dfrac{d}{{dx}}\tan \theta = {{\sec }^2}\theta } \right)$
So, by substituting these values, we get,
$\int {\dfrac{1}{{1 + {x^2}}}dx = \int {\dfrac{1}{{1 + {{\tan }^2}\theta }}} } ({\sec ^2}\theta )d\theta$
We know the identity ${\sec ^2}\theta - {\tan ^2}\theta = 1$ from which we get that $1 + {\tan ^2}\theta = {\sec ^2}\theta$ .
So,
$\int {\dfrac{1}{{1 + {{\tan }^2}\theta }}({{\sec }^2}\theta )d\theta = \int {\dfrac{{{{\sec }^2}\theta }}{{{{\sec }^2}\theta }}} } d\theta$
$\Rightarrow \int 1 d\theta = \theta$
And from the substitution that $x = \tan \theta$ , we get $\theta = {\tan ^{ - 1}}x$
So, we can conclude that, $\int {\dfrac{1}{{1 + {x^2}}}dx = {{\tan }^{ - 1}}x + c}$ ------(2)
So, from (1) and (2),
$\int {\dfrac{{{x^4}}}{{1 + {x^2}}}dx = \int {({x^2} - 1)dx + \int {\dfrac{1}{{1 + {x^2}}}dx = } } \dfrac{{{x^3}}}{3} - x + {{\tan }^{ - 1}}x + c}$
So, the value of integral is
$\int {\dfrac{{{x^4}}}{{1 + {x^2}}}dx = \dfrac{{{x^3}}}{3} - x + {{\tan }^{ - 1}}x + c}$ -----where “c” is an arbitrary constant.

Note:
After integrating and finding an indefinite integral, make sure that you add an arbitrary constant “c” to it. Also, remember the formula $\int {\dfrac{1}{{1 + {x^2}}}dx = {{\tan }^{ - 1}}x + c}$ which will be very useful to you. After finding partial fractions, once again simplify them and check whether you are getting the same result from which you found the partial fractions.