We #160;have,I= #8747; #160;3 #160;dx1 x #160;1+x2=3 #8747;dx1 x #160;1+x2Let #160;11 x #160;1+x2=A1 x+Bx+Cx2+1 #8658;11 x #160;x2+1=Ax2+1 ...

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Integration by Parts

∫β 1-x 1+x 2 ...

Question

$\int \frac{3}{(1 - x) (1 + x^{2})} d x$

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Solution

$We have, I = \int \frac{3 d x}{(1 - x) (1 + x^{2})} = 3 \int \frac{d x}{(1 - x) (1 + x^{2})} Let \frac{1}{(1 - x) (1 + x^{2})} = \frac{A}{1 - x} + \frac{B x + C}{x^{2} + 1} \Rightarrow \frac{1}{(1 - x) (x^{2} + 1)} = \frac{A (x^{2} + 1) + (B x + C) (1 - x)}{(1 - x) (x^{2} + 1)} \Rightarrow 1 = A x^{2} + A + B x - B x^{2} + C - C x \Rightarrow 1 = (A - B) x^{2} + (B - C) x + A + C Equating coefficients of like terms . A - B = 0 . . . . . (1) B - C = 0 . . . . . (2) A + C = 1 . . . . . (3) Solving (1), (2) and (3), we get A = \frac{1}{2}, B = \frac{1}{2}, C = \frac{1}{2} ∴ \frac{1}{(1 - x) (x^{2} + 1)} = \frac{1}{2 (1 - x)} + \frac{\frac{x}{2} + \frac{1}{2}}{x^{2} + 1} \int \frac{3 d x}{(1 - x) (x^{2} + 1)} = \frac{3}{2} \int \frac{d x}{1 - x} + \frac{3}{2} \int \frac{x d x}{x^{2} + 1} + \frac{3}{2} \int \frac{d x}{x^{2} + 1} Putting x^{2} + 1 = t \Rightarrow x d x = \frac{d t}{2} ∴ I = \frac{3}{2} \int \frac{d x}{1 - x} + \frac{3}{4} \int \frac{d t}{t} + \frac{3}{2} \int \frac{d x}{x^{2} + 1} = \frac{3}{2} \frac{\log |1 - x|}{- 1} + \frac{3}{4} \log |t| + \frac{3}{2} \times \tan^{- 1} x + C = \frac{- 3}{2} \log |1 - x| + \frac{3}{4} \log |1 + x^{2}| + \frac{3}{2} \tan^{- 1} x + C = \frac{- 3}{4} \times 2 \log |1 - x| + \frac{3}{4} \log |1 + x^{2}| + \frac{3}{4} (2 \tan^{- 1} x) + C = \frac{3}{4} [\log |1 + x^{2}| - \log |{(1 - x)}^{2}|] + \frac{3}{4} (2 \tan^{- 1} x) + C = \frac{3}{4} [\log |\frac{1 + x^{2}}{{(1 - x)}^{2}}| + 2 \tan^{- 1} (x)] + C$

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