Question

How do you integrate $\dfrac{{{x^2}}}{{{x^2} + 1}}$ ?

Answer 1

Hint : In this question, we have to find the integration of $\dfrac{{{x^2}}}{{{x^2} + 1}}$ so we must know what integration actually is. When we are given the differentiation of a function and we have to find the function, we integrate the differentiated function. We can get different values of integral by varying the value of the arbitrary constant so a function can have an infinite number of integrals but every function has a unique derivative. For solving such types of questions, we must know the integration of basic functions like.

Complete step-by-step answer :
We are given $\dfrac{{{x^2}}}{{{x^2} + 1}}$
It can be rewritten as –
$\dfrac{{{x^2} + 1 - 1}}{{{x^2} + 1}} \\\ \Rightarrow \dfrac{{{x^2}}}{{{x^2} + 1}} = 1 - \dfrac{1}{{{x^2} + 1}} \;$
Integrating the simplified expression, we get –
$\int {\dfrac{{{x^2}}}{{{x^2} + 1}}} dx = \int {(1 - \dfrac{1}{{{x^2} + 1}})dx} = \int {1dx} - \int {\dfrac{1}{{1 - {x^2}}}dx} \\\ \Rightarrow \int {\dfrac{{{x^2}}}{{{x^2} + 1}}} = x - \int {\dfrac{1}{{{x^2} + 1}}} dx \\\$
Now, let
$x = \tan \theta \, \Rightarrow \theta = {\tan ^{ - 1}}x \\\ \Rightarrow \dfrac{{dx}}{{d\theta }} = {\sec ^2}\theta \\\ \Rightarrow dx = {\sec ^2}\theta d\theta \;$
We know that
${\sec ^2}\theta - {\tan ^2}\theta = 1 \\\ {\tan ^2}\theta + 1 = {\sec ^2}\theta \\\ \Rightarrow {x^2} + 1 = {\sec ^2}\theta \;$
Using this in the obtained expression, we get –
$\int {\dfrac{{{x^2}}}{{{x^2} + 1}}dx} = x - \int {\dfrac{1}{{{{\sec }^2}\theta }} \times {{\sec }^2}\theta d\theta } \\\ \Rightarrow \int {\dfrac{{{x^2}}}{{{x^2} + 1}}dx} = x - \int {d\theta } \\\ \Rightarrow \int {\dfrac{{{x^2}}}{{{x^2} + 1}}dx} = x - \theta + c \\\ \Rightarrow \int {\dfrac{{{x^2}}}{{{x^2} + 1}}dx} = x - {\tan ^{ - 1}}x + c \\\$
Hence, the integration of $\dfrac{{{x^2}}}{{{x^2} + 1}}$ is $x - {\tan ^{ - 1}}x + c$
So, the correct answer is “ $x - {\tan ^{ - 1}}x + c$ ”.

Note : In this question, the function whose integration we have to find is a fraction containing polynomials in the numerator and the denominator but the answer came in the form of trigonometric ratios, so students must not get confused. We have used a trigonometric identity which states that the difference of the squares of the secant function and the tangent function is equal to 1 in this question. We also note that we need to memorize the differentiation and integration of some basic functions, so we remember that the differentiation of ${\tan ^{ - 1}}x$ is equal to $\dfrac{1}{{{x^2} + 1}}$ and we know that integration is the inverse process of differentiation, so the integration of $\dfrac{1}{{{x^2} + 1}}$ is equal to ${\tan ^{ - 1}}x$ . So, this question can also be solved by using this information.