Answer
$\int\frac{x}{(x^2+1)^{1.3}}dx=-\frac{5}{3(x^2+1)^{0.3}}+C$
Work Step by Step
Substitution:
$u=x^2+1$
$\frac{du}{dx}=2x$
$dx=\frac{1}{2x}du$
$\int\frac{x}{(x^2+1)^{1.3}}dx=\int\frac{x}{u^{1.3}}\frac{1}{2x}du=\int\frac{1}{2}u^{-1.3}du=\frac{1}{2}\frac{u^{-0.3}}{-0.3}+C=-\frac{1}{0.6}(x^2+1)^{-0.3}+C=-\frac{5}{3(x^2+1)^{0.3}}+C$
