\(\int\displaystyle\frac{2x}{\sqrt[3]{x^2 + 1}}\,dx\) is equal to (where C is a constant of integration)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{3}{2}(x^2 + 1)^{2/3} + C\)
Let \(t = x^2 + 1\). Then \(dt = 2x\,dx\), which is exactly the numerator.
\( \displaystyle \int \frac{2x}{\sqrt[3]{x^2 + 1}}\,dx = \int t^{-1/3}\,dt \)
\( \displaystyle = \frac{t^{2/3}}{2/3} + C \)
\( \displaystyle = \frac{3}{2}(x^2 + 1)^{2/3} + C \)