What is the integral of $int x^3/(1+x^2)dx$ ?

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Answer 1 · 2018-03-21T15:51:00

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

Perform the substitution

$u=1+x^2$ , $=>$ , $du=2xdx$

$=>$ , $x^2=u-1$

Therefore,

$int(x^3dx)/(1+x^2)=int(x^2xdx)/(1+x^2)=1/2int((u-1)du)/(u)$

$=1/2int1du-1/2int(du)/u$

$=1/2u-1/2lnu$

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

Answer 2 · 2018-03-21T16:02:04

$int x^3/(x^2+1)*dx=x^2/2-1/2Ln(x^2+1)+C$

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

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

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

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

= $x^2/2-1/2Ln(x^2+1)+C$

What is the integral of int x^3/(1+x^2)dxint x^3/(1+x^2)dx?