If d x ( x + 2 ) ( x 2 + 1 ) = a log | 1 + x 2 | + b tan - 1 x + 1 5 log | x + 2 | + C , then:

Question

If $\int \frac{d x}{(x + 2) (x^{2} + 1)} = a \log$ $| 1 + x^{2} |$ $+ b \tan^{- 1} x + \frac{1}{5} \log$ $| x + 2 |$ $+ C,$ then:

Smart Achievers · Accepted Answer

(3)

Ans. $a = \frac{- 1}{10}, b = \frac{2}{5}$

Explanation :

$Given that,$

$\int \frac{d x}{(x + 2) (x^{2} + 1)} = a \log$ $| 1 + x^{2} |$ $+ b \tan^{- 1} x + \frac{1}{5} \log$ $| x + 2 |$ $+ C$

$Now,$ $I = \int \frac{d x}{(x + 2) (x^{2} + 1)}$

$\frac{1}{(x + 2) (x^{2} + 1)} = \frac{A}{x + 2} + \frac{B x + C}{x^{2} + 1}$

$\Rightarrow$ $1 = A (x^{2} + 1) + (B x + C) (x + 2)$

$\Rightarrow$ $1 = A x^{2} + A + B x^{2} + 2 B x + C x + 2 C$

$\Rightarrow$ $1 = (A + B) x^{2} + (2 B + C) x + A + 2 C$

$\Rightarrow$ $A + B = 0,$ $A + 2 C = 1,$ $2 B + C = 0$

$We have, A = \frac{1}{5},$ $B = - \frac{1}{5}$ and $C = \frac{2}{5}$

$∴$ $\int \frac{d x}{(x + 2) (x^{2} + 1)} = \frac{1}{5} \int \frac{d x}{x + 2} + \int \frac{- \frac{1}{5} x + \frac{2}{5}}{x^{2} + 1} d x$

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

$= \frac{1}{5} \int \frac{d x}{x + 2} - \frac{1}{10} \int \frac{2 x}{1 + x^{2}} d x + \frac{2}{5} \int \frac{d x}{1 + x^{2}}$

$= \frac{1}{5} \log$ $| x + 2 |$ $- \frac{1}{10} \log$ $| 1 + x^{2} |$ $+ \frac{2}{5} \tan^{- 1} x + C$

$∴$ $b = \frac{2}{5}$ and $a = \frac{- 1}{10}$

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