Let $f$ be a continuous function satisfying
$f (x + y) = f (x) + f (y),$ for each $x, y \in R$
and $f (1) = 2$ then $\int \frac{f (x) \tan^{- 1} x}{{(1 + x^{2})}^{2}} d x$ is equal to

Moderate

A

cannot be determined explicitly
B

$C - \frac{\tan^{- 1} x}{2 (1 + x^{2})} + \frac{1}{4} \tan^{- 1} x + \frac{f (x)}{1 + (f (x))^{2}}$
C

$C - \frac{\tan x}{2 (1 + x^{2})} + \frac{\tan^{- 1} x}{4} + \frac{x}{4 (1 + x^{2})}$
D

$C - \frac{1}{(1 + x^{2})} \tan^{- 1} x + \frac{1}{2} \tan^{- 1} x + \frac{x}{2 (1 + x^{2})}$

Solution

First show that $f (x) = 2 x$ . Now ,
$I = \int \frac{f (x) \tan^{- 1} x}{{(1 + x^{2})}^{2}} d x = \int \frac{2 x}{{(1 + x^{2})}^{2}} \tan^{- 1} x d x$
Put $x = \tan θ$ , so that
$\begin{array}{l} I = \int \frac{2 \tan θ}{\sec^{4} θ} θ \sec^{2} θ d θ \\ = \int θ \sin (2 θ) d θ \\ = - \frac{1}{2} θ \cos 2 θ + \frac{1}{4} \sin 2 θ + C \\ = C - \frac{1}{2} θ \frac{1 - \tan^{2} θ}{1 + \tan^{2} θ} + \frac{1}{2} \frac{\tan θ}{1 + \tan^{2} θ} \\ = C - \frac{1}{2} \frac{1 - x^{2}}{1 + x^{2}} \tan^{- 1} x + \frac{1}{2} \frac{x}{1 + x^{2}} \end{array}$
$= C - \frac{1}{1 + x^{2}} \tan^{- 1} x + \frac{1}{2} \tan^{- 1} x + \frac{1}{2} \frac{x}{1 + x^{2}}$

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