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

Question

2165 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#