Find $\int \frac{2x}{\sqrt[3]{x^2 + 1}} dx$.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{3}{2} (x^2 + 1)^{2/3} + C$
$I=\int \frac{2x}{\sqrt[3]{x^2 + 1}} dx$
Let $x^2 + 1 = t$, then $2x dx = dt$.
$= \int \frac{dt}{t^{1/3}} = \int t^{-1/3} dt$
$= \frac{t^{2/3}}{2/3} + C = \frac{3}{2} t^{2/3} + C$
$= \frac{3(x^2 + 1)^{2/3}}{2} + C$