How do you evaluate $int x/(x^2+2)^3 dx$ for [-2, -1]?

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Answer 1 · 2015-10-15T15:04:44

$-33/3136$

Use substitution technique and let $u=x^2+3$
Then $du=2xdx$
$therefore 1/2du=xdx$

Limits : $x=-1=>u=4 and x=-2=>u=7$

We may hence re-write this integral as

$1/2int_4^7u^(-3)du=1/2[(u^(-2))/(-2)]_7^4$

$=-1/4(1/16-1/49)=-33/3136$

How do you evaluate int x/(x^2+2)^3 dxint x/(x^2+2)^3 dx for [-2, -1]?