Evaluate the following integrals:


Evaluate: $\int \frac{\mathrm{x}^{2}}{1+\mathrm{x}^{3}}$


let $1+x^{3}=t$

Differentiating on both sides we get,

$3 x^{2} d x=d t$

$x^{2} d x=\frac{1}{3} d t$

substituting it in $\int \frac{x^{2}}{1+x^{3}} d x$ we get,

$=\int \frac{1}{3 t} d t$

$=\frac{1}{3} \log t+c$

$=\frac{1}{3} \log \left(1+x^{3}\right)+c$

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