Question
Evaluate: $\int \frac{d x}{x^2+4 x+8}$
✓
Answer
Consider the integral, $\int \frac{d x}{x^2+4 x+8}$
$I =\int \frac{1}{x^2+4 x+8} dx$
$=\int \frac{1}{x^2+4 x+4+4} dx$
$=\int \frac{1}{(x+2)^2+2^2} d x \quad\{$ Use Intergral Formula :}
$\int \frac{1}{x^2+a^2} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)$
$=\frac{1}{2} \tan ^{-1} \frac{x+2}{2}+C$
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