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Question Number 141531 by sarkor last updated on 20/May/21

Commented by mohammad17 last updated on 20/May/21

$$let{z}^6=x+1\Rightarrow 6z^{5}dz=dx$$$$$$$$\int \frac{6z^3(z^4+z)}{(z+1)}dz=6\int \frac{z^7+z^4}{z+1)}dz$$$$$$$$=6\int z^4(z^2-z+1)dz=6(\frac{z^7}{7}-\frac{z^6}{6}+\frac{z^5}{5})+C$$$$$$$$set:z=\sqrt[6]{x+1}$$$$$$$$\therefore \int \frac{\sqrt[3]{{(x+1)}^2}+\sqrt[6]{x+1}}{\sqrt{x+1}+\sqrt[3]{x+1}}dx=\frac{6}{7}\sqrt[6]{{(x+1)}^7}-(x+1)+\frac{6}{5}\sqrt[6]{{(x+1)}^5}+C$$$$$$$$by:⟨m.o⟩$$$$$$

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