Solve -x2 - 1x2 - 1 dx

The correct option is C x 2 tan^ 1 x + c ∫x^2 1x^2+1dx =∫x^2+1 2x^2+1dx =∫ ( 1 2x^2+1 )dx =∫ 1dx ∫2x^2+1dx =x 2tan^ 1x+c ...

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Standard Formulae - 1

Solve ∫x2 -...

Question

Solve $\int \frac{x^{2} - 1}{x^{2} + 1} d x$

x - t a n^{- 1} x + c

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2 x - 2 t a n^{- 1} x + c

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x + 2 t a n^{- 1} x + c

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x - 2 t a n^{- 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

\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

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

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\int \frac{(x^{2} + 1) d x}{(x^{2} + 2) (2 x^{2} + 1)} = \frac{1}{3 \sqrt{(2)}} {{tan}^{- 1} \frac{x}{\sqrt{(2)}} + {tan}^{- 1} x \sqrt{2}} .

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

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