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Integration by Parts

∫ x x+1 x 2+1...

Question

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

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Solution

$We have, I = \int \frac{x d x}{(x + 1) (x^{2} + 1)} Let \frac{x}{(x + 1) (x^{2} + 1)} = \frac{A}{x + 1} + \frac{B x + C}{x^{2} + 1} \Rightarrow \frac{x}{(x + 1) (x^{2} + 1)} = \frac{A (x^{2} + 1) + (B x + C) (x + 1)}{(x + 1) (x^{2} + 1)} \Rightarrow x = A (x^{2} + 1) + B x^{2} + B x + C x + C \Rightarrow x = (A + B) x^{2} + (B + C) x + (A + C) Equating coefficients of like terms A + B = 0 . . . . . (1) B + C = 1 . . . . . (2) A + C = 0 . . . . . (3) Solving (1), (2) and (3), we get A = - \frac{1}{2} B = \frac{1}{2} C = \frac{1}{2} ∴ \frac{x}{(x + 1) (x^{2} + 1)} = - \frac{1}{2 (x + 1)} + \frac{\frac{x}{2} + \frac{1}{2}}{x^{2} + 1} \Rightarrow \int \frac{x d x}{(x + 1) (x^{2} + 1)} = - \frac{1}{2} \int \frac{d x}{x + 1} + \frac{1}{2} \int \frac{x d x}{x^{2} + 1} + \frac{1}{2} \int \frac{d x}{x^{2} + 1} let x^{2} + 1 = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = - \frac{1}{2} \int \frac{d x}{x + 1} + \frac{1}{4} \int \frac{d t}{t} + \frac{1}{2} \int \frac{d x}{x^{2} + 1^{2}} = - \frac{1}{2} \log |x + 1| + \frac{1}{4} \log |t| + \frac{1}{2} \tan^{- 1} x + C' = - \frac{1}{2} \log |x + 1| + \frac{1}{4} \log |x^{2} + 1| + \frac{1}{2} \tan^{- 1} x + C'$

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