The correct option is C log|x+1 |+tan ^ 1x + c ∫(x 1)/(x+1)(x^2+1)dx (x 1) / (x+1)(x^ 2 +1) = A / (x+1) + Bx+C / (x^ 2 +1) ⇒ x 1=A(x^ 2 ...

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Functions

∫x-1/x+1x2+1d...

Question

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

\frac{1}{2} log | \frac{x + 1}{\sqrt{x^{2} + 1}} | + c

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- log | x + 1 | + {tan}^{- 1} x + c

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log | \frac{\sqrt{x^{2} + 1}}{x + 1} | + c

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log | x + 1 | + \frac{1}{2} {tan}^{- 1} x + c

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Solution

The correct option is C $- log | x + 1 | + {tan}^{- 1} x + c$
$\int \frac{(x - 1)}{(x + 1) (x^{2} + 1)} d x$
$\frac{(x - 1)}{(x + 1) (x^{2} + 1)} = \frac{A}{(x + 1)} + \frac{B x + C}{(x^{2} + 1)}$
$\Rightarrow x - 1 = A (x^{2} + 1) + (B x + C) (x + 1)$
$x - 1 = (B + C) x + (A + B) x^{2} + (A + C)$
$\Rightarrow B + C = 1, A + B = 0, A + C = - 1$
Solving these equations, we get
$C = 0, A = - 1, B = 1$
$\frac{(x - 1)}{(x + 1) (x^{2} + 1)} = \frac{- 1}{(x + 1)} + \frac{1}{(x^{2} + 1)}$
Integrating w.r.t x
$\int \frac{(x - 1)}{(x + 1) (x^{2} + 1)} d x =$ $- log | x + 1 | + {tan}^{- 1} x + c$

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