Let #160;I= #8747;x+1x2+4x+5dx #38; #160;let #160;x+1= #160;Addxx2+4x+5+B #8658;x+1=A #160;2x+4+B #8658;x+1=2Ax+4A+BEquating #160;the #160 ...

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Factorization

∫ x+1 x 2+4 x...

Question

$\int \frac{x + 1}{x^{2} + 4 x + 5} d x$

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Solution

$Let I = \int \frac{(x + 1)}{x^{2} + 4 x + 5} d x & let (x + 1) = A \frac{d}{d x} (x^{2} + 4 x + 5) + B \Rightarrow x + 1 = A (2 x + 4) + B \Rightarrow x + 1 = (2 A) x + 4 A + B Equating the coefficients of like terms 2 A = 1 \Rightarrow A = \frac{1}{2} & 4 A + B = 1 \Rightarrow 4 \times \frac{1}{2} + B = 1 \Rightarrow B = - 1 ∴ (x + 1) = \frac{1}{2} (2 x + 4) - 1 ∴ I = \int [\frac{\frac{1}{2} (2 x + 4) - 1}{x^{2} + 4 x + 5}] d x = \frac{1}{2} \int \frac{(2 x + 4)}{x^{2} + 4 x + 5} d x - \int \frac{1}{x^{2} + 4 x + 5} d x Putting x^{2} + 4 x + 5 = t \Rightarrow (2 x + 4) d x = d t ∴ I = \frac{1}{2} \int \frac{1}{t} d t - \int \frac{1}{x^{2} + 4 x + 4 + 1} d x = \frac{1}{2} \int \frac{d t}{t} - \int \frac{1}{{(x + 2)}^{2} + 1^{2}} d x = \frac{1}{2} \ln |t| - \tan^{- 1} (\frac{x + 2}{1}) + C [∵ \int \frac{1}{x^{2} + a^{2}} d x = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C] = \frac{1}{2} \ln |x^{2} + 4 x + 5| - \tan^{- 1} (x + 2) + C [∵ t = x^{2} + 4 x + 5]$

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