$\int \frac{d x}{x^{2} + 2 x + 2}$ equals:
$(A) x {tan}^{- 1} (x + 1) + C$
$(B) {tan}^{- 1} (x + 1) + C$
$(C) (x + 1) {tan}^{- 1} x + C$
$(D) {tan}^{- 1} x + C$

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Solution

$\int \frac{d x}{x^{2} + 2 x + 2}$
$= \int \frac{d x}{x^{2} + 2 x + 1 + 1}$
$= \int \frac{d x}{(x + 1)^{2} + (1)^{2}}$
$= \frac{1}{1} {tan}^{- 1} \frac{x + 1}{1} + C$
$[∵ \int \frac{1}{x^{2} + a^{2}} d x = \frac{1}{a} {tan}^{- 1} \frac{x}{a} + C]$
$= {tan}^{- 1} (x + 1) + C$
Where C is constant of integration
$∴$ Option B is correct.

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