Answer
$\int\frac{x^2}{(1+x^3)^{1.4}}dx=-\frac{5}{6(1+x^3)^{0.4}}+C$
Work Step by Step
Substitution:
$u=1+x^3$
$\frac{du}{dx}=3x^2$
$dx=\frac{1}{3x^2}du$
$\int\frac{x^2}{(1+x^3)^{1.4}}dx=\int\frac{x^2}{u^{1.4}}\frac{1}{3x^2}du=\int\frac{1}{3}u^{-1.4}du=\frac{1}{3}\frac{u^{-0.4}}{-0.4}+C=-\frac{1}{1.2}(1+x^3)^{-0.4}+C=-\frac{5}{6(1+x^3)^{0.4}}+C$
