How do you integrate $int x(x^2+1)^3dx$ from [-1,1]?

Question

2429 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-09T12:42:20

We know that,
If function $f$ is odd and continous in $[-a,a]$ , then

$color(red)(int_-a^af(x)dx=0,ainR^+$

Here,

$I=int_-1^1 x(x^2+1)^3dx$

Please see the graph of $y=x(x^2+1)^3$
graph{x(x^2+1)^3 [-12.3, 12.3, -6.135, 6.175]}

Let, $f(x)=x(x^2+1)^3$

$f(-x)=(-x)((-x)^2+1)^3=-x(x^2+1)^3=-f(x)$

So,

$f(-x)=-f(x)=>f$ is odd function,

and from the above graph we can say that

$f$ is continous on [-1,1].

Hence,

$int_-1^1x(x^2+1)^3dx=0$

How do you integrate int x(x^2+1)^3dxint x(x^2+1)^3dx from [-1,1]?