Question

Evaluate the integral
∫▒√(〖1+x〗^(-2/3) ) dx
using the substitution u=1+x^(-2/3) dx

Answer 1

$\int\sqrt{(1+x)^{-2/3}}dx=\int({1+x})^{-1/3}dx \\ u=x+1 \implies \frac{du}{dx}=1\implies\ du=dx\implies\int\frac{dx}{\sqrt[3]{1+x}}= \int\frac{du}{\sqrt[3]{u}}=\frac{3}{2}u^{2/3}+C=\frac{3}{2}(x+1)^{2/3}+C$