∫x2+2(x2+1)(x2+4)dx=Atan−1x+Btan−1x2+c then (A, B) =

∫x2+2(x2+1)(x2+4)dx=∫13x2+1+23x2+4dx=13∫1x2+1dx+23∫1x2+22dx=13tan-1x+2312tan-1x2+C∴∫1x2+a2dx=1atan−1xa∴A=13, B=13.

∫x2+2(x2+1)(x2+4)dx=Atan−1x+Btan−1x2+c then (A, B) =

∫x2+2(x2+1)(x2+4)dx=∫13x2+1+23x2+4dx=13∫1x2+1dx+23∫1x2+22dx=13tan-1x+2312tan-1x2+C∴∫1x2+a2dx=1atan−1xa∴A=13, B=13.

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$\int \frac{x^{2} + 2}{(x^{2} + 1) (x^{2} + 4)} d x = A \tan^{- 1} x + B \tan^{- 1} \frac{x}{2} + c$ then (A, B) =

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$(\frac{1}{2}, \frac{1}{3})$

$(\frac{- 1}{2}, \frac{- 2}{3})$

$(\frac{1}{4}, \frac{1}{3})$

$(\frac{1}{3}, \frac{1}{3})$

answer is D.

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Detailed Solution

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

$= \int [\frac{\frac{1}{3}}{x^{2} + 1} + \frac{\frac{2}{3}}{x^{2} + 4}] d x$

$= \frac{1}{3} \int \frac{1}{x^{2} + 1} d x + \frac{2}{3} \int \frac{1}{x^{2} + 2^{2}} d x$

$= \frac{1}{3} (\tan^{- 1} x) + \frac{2}{3} \frac{1}{2} \tan^{- 1} \frac{x}{2} + C$

$(∴ \int \frac{1}{x^{2} + a^{2}} d x = \frac{1}{a} \tan^{- 1} \frac{x}{a})$

$∴ A = \frac{1}{3}, B = \frac{1}{3} .$

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∫x2+2(x2+1)(x2+4)dx=Atan−1x+Btan−1x2+c then (A, B) =

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$\int \frac{x^{2} + 2}{(x^{2} + 1) (x^{2} + 4)} d x = A \tan^{- 1} x + B \tan^{- 1} \frac{x}{2} + c$ then (A, B) =