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Question Number 68095 by mhmd last updated on 04/Sep/19

$$\int \sqrt{x}/1+\sqrt[3]{x}dx$$

Answered by MJS last updated on 05/Sep/19

$$\int \frac{x^{1/2}}{1+x^{1/3}}dx=$$$$[t=x^{1/6}\to dx=6x^{5/6}dt]$$$$=6\int \frac{t^8}{t^2+1}dt=6\int (t^6-t^4+t^2-1+\frac{1}{t^2+1})dt=$$$$=\frac{6}{7}{t}^7-\frac{6}{5}{t}^5+2t^3-6t+6\arctan t=$$$$=\frac{6}{7}{x}^{7/6}-\frac{6}{5}{x}^{5/6}+2x^{1/2}-6x^{1/6}+6\arctan x^{1/6}+C$$

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