Answer:
[tex]\frac{1}{6} (arctan(x^3))^2 +C[/tex]
Step-by-step explanation:
We are given
[tex]\int {\frac{x^2arctan(x^3)}{1+x^6} } \, dx[/tex]
First, we can apply a u substitution where [tex]u=x^3\\du=3x^2 dx[/tex]
To get our du, we will need to add a 3 to the top. That will give us
[tex]\frac{1}{3} \int {\frac{3x^2arctan(x^3)}{1+x^6} } \, dx[/tex]
Now we can apply our substitution
[tex]\frac{1}{3} \int {\frac{arctan(u)}{1+u^2} } \, du[/tex]
Next, we need another substitution. This time, lets use the variable v where [tex]v=arctan(u)\\dv=\frac{1}{1+u^2} du[/tex]
We already have our dv, so we can apply this substitution as well
[tex]\frac{1}{3} \int v \,dv\\\\\frac{1}{6} v^2 +C[/tex]
Now we need to get back to x, so we will have to sub in v and then u to do so
[tex]\frac{1}{6} (arctan(u))^2 +C\\\\\frac{1}{6} (arctan(x^3))^2 +C[/tex]
