Evaluate: 8 (x+2) (x^2+4 ) d x - Vidyadip

Question

Evaluate: $\int \frac{8}{(x+2)\left(x^2+4\right)} d x$

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Answer

$ \text { Let } I=\int \frac{8}{(x+2)\left(x^2+4\right)} d x$
$ \text { Let } \frac{8}{(x+2)\left(x^2+4\right)}=\frac{A}{x+2}+\frac{B x+C}{x^2+4}$
$ 8=A\left(x^2+4\right)+(B x+C)(x+2)$
$ 8=A\left(x^2+4\right)+B x^2+2 B x+C x+2 C$
$ 8=(A+B) x^2+(2 B+c) x+(4 A+2 C)$
Comparing the coefficients of $x^2 , x$ and the constant term, we get
$A + B = 0, 2B + C = 0$ and $4A + 2C = 8$
On solving these equations, we get
$A = 1, B = –1, C = 2$
$ \frac{8}{(x+2)\left(x^2+4\right)}=\frac{1}{x+2}+\frac{-x+2}{x^2+4}$
$ I=\int\left[\frac{1}{x+2}+\frac{-x+2}{x^2+4}\right] d x$
$ =\int \frac{1}{x+2} d x-\frac{1}{2} \int \frac{2 x}{x^2+4} d x+2 \int \frac{1}{x^2+2^2} d x$
$ =\log |x+2|-\frac{1}{2} \log \left|x^2+4\right|+\tan ^{-1}\left(\frac{x}{2}\right)+c$
$ =\log \left|\frac{x+2}{\sqrt{x^2+4}}\right|+\tan ^{-1}\left(\frac{x}{2}\right)+c$

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