How do you integrate $int x^3/(x^2+2)^3$ by integration by parts method?

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Answer 1 · 2017-01-23T02:47:12

Integration by parts is: $intudv=uv-intvdu$
choose $u$ so that $dv$ can be integrated by variable substitution.

let $u=x^2 and dv=(x)/(x^2+2)^3dx$ , because this will allow:

$v = intx/(x^2+2)^3dx$

to be integrated by letting $t = x^2+2$ , then $dt = 2xdx$ or $dt/2 = xdx$

This makes the integral become:

$v = 1/2intt^-3dt$

$v = -1/4t^-2$

Reversing the substitution:

$v=-1/(4(x^2+2)^2$

and $du=2xdx$

$intx^3/(x^2+2)^3dx = (x^2)(-1/(4(x^2+2)^2)) - int-1/(x^2+2)^2(2x)dx$

$intx^3/(x^2+2)^3dx = -x^2/(4(x^2+2)^2) + 1/4int(2x)/(x^2+2)^2dx$

The last integral is the same sort of variable substitution:

$intx^3/(x^2+2)^3dx = -x^2/(4(x^2+2)^2) - 1/(4(x^2+2)) + C$

How do you integrate int x^3/(x^2+2)^3int x^3/(x^2+2)^3 by integration by parts method?