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Question Number 33690 by mondodotto@gmail.com last updated on 22/Apr/18

$$\int \frac{\mathbf{dx}}{\sqrt[3]{\mathbf{x}}+\sqrt{\mathbf{x}}}$$

Answered by math1967 last updated on 22/Apr/18

$$letx=z^6\therefore dx=6z^{5}dz$$$$6\int \frac{z^{5}dz}{z^2+z^3}=6\int \frac{z^{5}dz}{z^2(1+z)}=6\int \frac{z^{3}dz}{1+z}$$$$=6\int \frac{(z^3+1)dz}{z+1}-6\int \frac{dz}{z+1}$$$$=6\int (z^2-z+1)dz-6ln[z+1]$$$$=6\frac{z^3}{3}-\frac{6z^2}{2}+6z-6ln[z+1]+c$$$$=2\sqrt{x}-3x^{\frac{1}{3}}+6x^{\frac{1}{6}}-6ln[x^{\frac{1}{6}}+1]+c$$

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