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Integrals of Mixed Powers of Sine and Cosine

∫ x 2 x 2+6 x...

Question

$\int \frac{x^{2}}{x^{2} + 6 x + 12} d x$

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Solution

$Let I = \int \frac{x^{2} d x}{x^{2} + 6 x + 12} Now, x^{2} + 6 x + 12 \overset{1}{x^{2}} x^{2} + 6 x + 12 - - - - 6 x - 12 Therefore, \frac{x^{2}}{x^{2} + 6 x + 12} = 1 - \frac{(6 x + 12)}{x^{2} + 6 x + 12} . . . . . (1) Let 6 x + 12 = A \frac{d}{d x} (x^{2} + 6 x + 12) + B \Rightarrow 6 x + 12 = A (2 x + 6) + B \Rightarrow 6 x + 12 = (2 A) x + 6 A + B Equating Coefficients of like terms 2 A = 6 A = 3 6 A + B = 12 18 + B = 12 B = - 6 ∴ \frac{x^{2}}{x^{2} + 6 x + 12} = 1 - \frac{3 (2 x + 6) - 6}{x^{2} + 6 x + 12} I = \int \frac{x^{2} d x}{x^{2} + 6 x + 12} = \int d x - 3 \int \frac{(2 x + 6) d x}{x^{2} + 6 x + 12} + 6 \int \frac{d x}{x^{2} + 6 x + 12} = \int d x - 3 \int \frac{(2 x + 6) d x}{x^{2} + 6 x + 12} + 6 \int \frac{d x}{x^{2} + 6 x + 9 + 3} = \int d x - 3 \int \frac{(2 x + 6) d x}{x^{2} + 6 x + 12} + 6 \int \frac{d x}{{(x + 3)}^{2} + {(\sqrt{3})}^{2}} = x - 3 \log |x^{2} + 6 x + 12| + \frac{6}{\sqrt{3}} \tan^{- 1} (\frac{x + 3}{\sqrt{3}}) + C = x - 3 \log |x^{2} + 6 x + 12| + 2 \sqrt{3} \tan^{- 1} (\frac{x + 3}{\sqrt{3}}) + C$

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