I have a problem with this integral:
$\int{x^3\over x^8+3}dx$
I try to substitute $x^4$, but I have no idea how continue. Thank you for any help.
I have a problem with this integral:
$\int{x^3\over x^8+3}dx$
I try to substitute $x^4$, but I have no idea how continue. Thank you for any help.
Let $t=x^4$, then $dt=4x^3\,dx$, and
$\int\frac{x^3}{x^8+3}\,dx=\int\frac{dt}{4(t^2+3)}$
Let $t=\sqrt3u$, then $dt=\sqrt3\,du$, and
$\int\frac{dt}{4(t^2+3)}=\int\frac{\sqrt3\,du}{12(u^2+1)}=\frac{\sqrt3}{12}\int\frac{du}{1+u^2}=\frac{\sqrt3}{12}\arctan u+C=\frac{\sqrt3}{12}\arctan\frac{\sqrt3x^4}{3}+C$