Question

what is the integration of (square root of x)/(1+(cube root of x))

Answer 1

Answer to Question #50467

$$\begin{array}{l} \int \frac {\sqrt {x}}{1 + \sqrt [ 3 ]{x}} d x = \left\{\text{substitute } u = \sqrt [ 6 ]{x} \text{ and } du = \frac {d x}{6 x ^ {\frac {5}{6}}} \right\} \\ = 6 \int \frac {u ^ {8}}{1 + u ^ {2}} d u \\ = 6 \int \left(u ^ {6} - u ^ {4} + u ^ {2} + \frac {1}{u ^ {2} + 1} - 1\right) d u \\ = 6 \left(\int \frac {1}{u ^ {2} + 1} d u + \int u ^ {6} d u - \int u ^ {4} d u + \int u ^ {2} d u - \int 1 \cdot d u\right) \\ = 6 \tan^ {- 1} u + \frac {6 u ^ {7}}{7} - \frac {6 u ^ {5}}{5} + 2 u ^ {2} - 6 u = \left\{\text{substitute back } u = \sqrt [ 6 ]{x} \right\} \\ = \frac {6 x ^ {\frac {7}{6}}}{7} - \frac {6 x ^ {\frac {5}{6}}}{5} + 2 \sqrt {x} - 6 \sqrt [ 6 ]{x} + 6 \tan^ {- 1} \left(\sqrt [ 6 ]{x}\right) + C \\ \end{array}$$

Answer:

$$\frac {6 x ^ {\frac {7}{6}}}{7} - \frac {6 x ^ {\frac {5}{6}}}{5} + 2 \sqrt {x} - 6 \sqrt [ 6 ]{x} + 6 \tan^ {- 1} \left(\sqrt [ 6 ]{x}\right) + C$$

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