First slide

Introduction to integration

Question

If $\int \frac{(x - 1)^{2}}{{(x^{2} + 1)}^{2}} d x = \tan^{- 1} x + g (x) + K$ then
$g (x)$ is equal to

Easy

A

$\tan^{- 1} (x / 2)$
B

$\frac{1}{x^{2} + 1}$
C

$\frac{1}{2 (x^{2} + 1)}$
D

none of these

Solution

$\begin{array}{l} \frac{(x - 1)^{2}}{{(x^{2} + 1)}^{2}} = \frac{x^{2} - 2 x + 1}{{(x^{2} + 1)}^{2}} \\ = \frac{x^{2} + 1}{{(x^{2} + 1)}^{2}} - \frac{2 x}{{(x^{2} + 1)}^{2}} \\ = \frac{1}{x^{2} + 1} - \frac{2 x}{{(x^{2} + 1)}^{2}} \end{array}$
so $\begin{array}{l} \int \frac{(x - 1)^{2}}{{(x^{2} + 1)}^{2}} d x = \int \frac{d x}{x^{2} + 1} - \int \frac{2 x}{{(x^{2} + 1)}^{2}} d x \\ = \tan^{- 1} x + \frac{1}{x^{2} + 1} + K \end{array}$
Therefore, $g (x) = \frac{1}{x^{2} + 1}$

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