$\int \frac{x^{4}}{(x - 1) (x^{2} + 1)} d x$

x^{2} + x + \frac{1}{2} l o g (x - 1) - \frac{1}{4} l o g (x^{2} + 1) - \frac{1}{2} {tan}^{- 1} x + c

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\frac{1}{2} x^{2} + x + \frac{1}{2} l o g (x - 1) - \frac{1}{4} l o g (x^{2} + 1) - \frac{1}{2} {tan}^{- 1} x + c

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\frac{1}{2} x^{2} - \frac{1}{2} l o g (x - 1) - \frac{1}{4} l o g (x^{2} + 1) - \frac{1}{2} {tan}^{- 1} x + c

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\frac{1}{2} x^{2} + \frac{1}{2} l o g (x - 1) - \frac{1}{4} l o g (x^{2} + 1) - \frac{1}{2} {tan}^{- 1} x + c

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Solution

The correct option is B $\frac{1}{2} x^{2} + x + \frac{1}{2} l o g (x - 1) - \frac{1}{4} l o g (x^{2} + 1) - \frac{1}{2} {tan}^{- 1} x + c$

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Similar questions

$\int \frac{x^{2} + 1}{x^{4} + 1} d x$ will be equal to which of the following

(1) {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 {tan}^{- 1} x, | x | \leq 1

(2) {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = 2 {tan}^{- 1} x, x \geq 0

(3) {tan}^{- 1} (\frac{2 x}{1 - x^{2}}) = 2 {tan}^{- 1} x, - 1 < x < 1

{cot}^{- 1} (\frac{\sqrt{1 + x^{2}} - 1}{x}) =

{tan}^{- 1} \frac{\sqrt{1 + x^{2}} - 1}{x} = \frac{1}{2} {tan}^{- 1} x, x \neq 0

\int \frac{e^{log (1 + 1 / x^{2})}}{x^{2} + 1 / x^{2}} d x = \frac{1}{\sqrt{(2)}} {tan}^{- 1} (x - \frac{1}{x})

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