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

Moderate

A

$\frac{1}{5} \log |x - 1| - \frac{1}{10} \log (x^{2} + 4) - \frac{2}{5} \tan^{- 1} \frac{x}{2} + C$
B

$\frac{1}{5} \log |x - 1| - \frac{1}{10} \log (x^{2} + 4) + \frac{4}{5} \tan^{- 1} \frac{x}{2} + C$
C

$\frac{1}{5} \log |x - 1| - \frac{1}{10} \log (x^{2} + 4) + \frac{2}{5} \tan^{- 1} \frac{x}{2} + C$
D

None of these

Solution

$\int \frac{x}{(x - 1) (x^{2} + 4)} d x l e t \frac{x}{(x - 1) (x^{2} + 4)} = \frac{A}{x - 1} + \frac{B x + C}{x^{2} + 4} - - - - - - - - (1) o r x = A (x^{2} + 4) + (B x + C) (x - 1) - - - - - - - (2) p u t t i n g x = 1 i n e q u a t i o n (2), w e g e t 1 = 5 A o r A = \frac{1}{5} p u t t i n g x = 0 i n e q u a t i o n (2), w e g e t 0 = 4 A - C o r C = 4 A = \frac{4}{5} p u t t i n g x = - 1 i n e q u a t i o n (2), w e g e t - 1 = 5 A + 2 B - 2 C s o l v i n g t h e s e e q u a t i o n s, w e g e t A = \frac{1}{5}, B = \frac{- 1}{5} a n d C = \frac{4}{5} S u b s t i t u t i n g t h e v a l u e s o f A, B a n d C i n e q u a t i o n (1), w e o b t a i n \frac{x}{(x - 1) (x^{2} + 4)} = \frac{1}{5 (x - 1)} + \frac{\frac{- 1}{5} x + \frac{4}{5}}{x^{2} + 4} = \frac{1}{5 (x - 1)} - \frac{1}{5} \frac{(x - 4)}{(x^{2} + 4)} I = \frac{1}{5} \int \frac{1}{(x - 1)} d x - \frac{1}{5} \int \frac{x - 4}{x^{2} + 4} d x = \frac{1}{5} \int \frac{1}{(x - 1)} d x - \frac{1}{10} \int \frac{2 x}{x^{2} + 4} d x + \frac{4}{5} \int \frac{1}{x^{2} + 4} d x = \frac{1}{5} \log |x - 1| - \frac{1}{10} \log (x^{2} + 4) + \frac{4}{5} \times \frac{1}{2} \tan^{- 1} \frac{x}{2} + C \frac{1}{5} \log |x - 1| - \frac{1}{10} \log (x^{2} + 4) + \frac{2}{5} \tan^{- 1} \frac{x}{2} + C$

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