CUET Preparation Today
Find $\int \frac{2x}{\sqrt[3]{x^2 + 1}} dx$. |
$\frac{2}{3} (x^2 + 1)^{3/2} + C$ $\frac{3}{2} (x^2 + 1)^{2/3} + C$ $3 (x^2 + 1)^{2/3} + C$ $\frac{3}{2} (x^2 + 1)^{1/3} + C$ |
$\frac{3}{2} (x^2 + 1)^{2/3} + C$ |
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 $2x dx = dt$. $= \int \frac{dt}{t^{1/3}} = \int t^{-1/3} dt$ $= \frac{t^{2/3}}{2/3} + C = \frac{3}{2} t^{2/3} + C$ $= \frac{3(x^2 + 1)^{2/3}}{2} + C$ |
