The correct option is B xlog ( 1+x^2 ) 2x+2tan^ 1x. I=∫log ( 1+x^2 )dx= ∫ 1.log ( 1+x^2 )dx xlog ( 1+x^2 ) ∫ x.1/1+x^2.2xdx xlog ( 1+x^2 ) ...

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Integration by Partial Fractions

Solve ∫log ...

Question

Solve $\int log (1 + x^{2}) d x$

x log (1 + x^{2}) - 2 x - 2 {tan}^{- 1} x .

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x log (1 + x^{2}) - 2 x + 2 {tan}^{- 1} x .

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x log (1 + x^{2}) + 2 x + 2 {tan}^{- 1} x .

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x log (1 - x^{2}) - 2 x + 2 {tan}^{- 1} x .

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Solution

The correct option is B $x log (1 + x^{2}) - 2 x + 2 {tan}^{- 1} x .$
$I = \int log (1 + x^{2}) d x = \int 1. log (1 + x^{2}) d x$
$x log (1 + x^{2}) - \int x . \frac{1}{1 + x^{2}} .2 x d x$
$x log (1 + x^{2}) - 2 \int \frac{x^{2} + 1 - 1}{x^{2} + 1} d x$
$= x log (1 + x^{2}) - 2 \int [1 - \frac{1}{1 + x^{2}}] d x$
$= x log (1 + x^{2}) - 2 x + 2 {tan}^{- 1} x .$

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Similar questions

(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

\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}

3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0

\tan^{- 1}

\sin

x

=

\tan

\sin

x

x \neq \frac{π}{2}

\cos^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) + \frac{1}{2} \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{2 π}{3}

\tan^{- 1} (\frac{x - 2}{x - 1}) + \tan^{- 1} (\frac{x + 2}{x + 1}) = \frac{π}{4}

{sin}^{- 1} \frac{2 x}{1 + x^{2}}; {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}; {tan}^{- 1} \frac{2 x}{1 - x^{2}} .

2 {tan}^{- 1} x .

\int 2 {tan}^{- 1} x = 2 [x {tan}^{- 1} x - \frac{1}{2} log (1 + x^{2})]

\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})

\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}

3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0

\tan^{- 1} (\frac{x - 2}{x - 1}) + \tan^{- 1} (\frac{x + 2}{x + 1}) = \frac{π}{4}

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