Evaluate: _ -4 ^2 1 x^2+4 x+13 d x - Vidyadip

Question

Evaluate: $\int_{-4}^2 \frac{1}{x^2+4 x+13} d x$

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Answer

$\text { Let } I =\int_{-4}^2 \frac{1}{x^2+4 x+13} d x$
$=\int_{-4}^2 \frac{1}{x^2+4 x+4+9} d x$
$=\int_{-4}^2 \frac{1}{(x+2)^2+(3)^2} d x$
$=\left[\frac{1}{3} \tan ^{-1}\left(\frac{x+2}{3}\right)\right]_{-4}^2$
$=\frac{1}{3}\left[\tan ^{-1}\left(\frac{4}{3}\right)-\tan ^{-1}\left(-\frac{2}{3}\right)\right]$
$\therefore I =\frac{1}{3}\left[\tan ^{-1}\left(\frac{4}{3}\right)+\tan -1\left(\frac{2}{3}\right)\right] \ldots \ldots .\left[\because \tan ^{-1}(-\theta)=-\tan ^{-1} \theta\right]$

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