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Question Number 53782 by maxmathsup by imad last updated on 25/Jan/19

$$calculate{\int }_0^1\frac{t^2}{1+t^3}dt$$

Answered by Smail last updated on 26/Jan/19

$${\int }_0^1\frac{t^2}{1+t^3}dt=\frac{1}{3}{\int }_0^1\frac{d(1+t^3)}{1+t^3}=\frac{1}{3}{[ln\mid 1+t^3\mid ]}_0^1$$$$=\frac{ln2}{3}$$

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