Question #fbe84

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Answer 1 · 2017-03-18T18:28:44

The answer is $=1/2ln(x^2+1)+1/(2x^2+2)+C$

We perform this integration by substitution

Let $u=x^2+1$

$du=2xdx$

$xdx=(du)/2$

Therefore,

$int(x^3dx)/(x^2+1)^2=int(x^2*xdx)/(x^+1)^2$

$=1/2int((u-1)du)/u^2$

$=1/2int(1/u-1/u^2)du$

$=1/2int(du)/u-1/2int(du)/u^2$

$=1/2lnu+1/2*1/u$

$=1/2ln(x^2+1)+1/(2x^2+2)+C$