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.