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Question Number 108663 by bobhans last updated on 18/Aug/20

Answered by john santu last updated on 18/Aug/20

$$\frac{\frac{⊸J}{S\leftrightharpoons }}{}$$$$let\sqrt{1+x^3}=m\Rightarrow x^3=m^2-1$$$$\Rightarrow 3x^{2}dx=2mdm$$$$I=\int \frac{x^3.x^2}{\sqrt{1+x^3}}dx=\frac{2}{3}\int \frac{(m^2-1)mdm}{m}$$$$=\frac{2}{3}\int (m^2-1)dm$$$$=\frac{2}{9}{m}^3-\frac{2}{3}m+C$$$$=\frac{2}{9}(1+x^3)-\frac{2}{3}\sqrt{1+x^3}+C$$

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