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CBSE Class 12 Math 2013 Solved Paper

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Question : 18 of 29
Marks: +1, -0
Evaluate: ∫ x2(x2+4)(x2+9)\frac{x^2}{(x^2+4)(x^2+9)} dx
Solution:  
Let x2x^2 = y
x2(x2+4)(x2+9)\frac{x^2}{(x^2+4)(x^2+9)} = y(y+)(y+9)\frac{y}{(y+)(y+9)} = Ay+4\frac{A}{y+4} + By+9\frac{B}{y+9}
y = A (y + 9) + B (y + 4)
Comparing both sides,
A + B = 1 and 9A + 4B = 0
Solving, we get A = −45\frac{-4}{5} and B = 95\frac{9}{5}
∴ I = ∫ ∣−45(x2+4)∣+95(x2+9)\left|\frac{-4}{5(x^2+4)}\right|+\frac{9}{5(x^2+9)} dx
= −45×12tan⁡−1(x2)-\frac{4}{5} \times \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + 95×13tan⁡−1(x3)\frac{9}{5} \times \frac{1}{3} \tan^{-1}\left(\frac{x}{3}\right) + X
= - 25tan⁡−1(x2)+35tan⁡−1(x3)\frac{2}{5} \tan^{-1}\left(\frac{x}{2}\right) + \frac{3}{5} \tan^{-1}\left(\frac{x}{3}\right) + C
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