$\int \frac{x^{2}}{1 + (x^{3})^{2}} d x$ is equal to

{tan}^{- 1} x^{2} + C

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\frac{2}{3} {tan}^{- 1} x^{3} + C

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\frac{1}{3} {tan}^{- 1} x^{3} + C

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\frac{2}{3} {tan}^{- 1} x^{2} + C

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Solution

The correct option is B $\frac{1}{3} {tan}^{- 1} x^{3} + C$
Let $I = \int \frac{x^{2}}{1 + (x^{3})^{2}} d x$

Put $x^{3} = t$

$∴ 3 x^{2} d x = d t$

$∴ I = \frac{1}{3} \int \frac{d t}{1 + t^{2}} = \frac{1}{3} {tan}^{- 1} t + C$

$= \frac{1}{3} {tan}^{- 1} (x^{3}) + C$

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