Answer to Question #50467
∫ x 1 + x 3 d x = { substitute u = x 6 and d u = d x 6 x 5 6 } = 6 ∫ u 8 1 + u 2 d u = 6 ∫ ( u 6 − u 4 + u 2 + 1 u 2 + 1 − 1 ) d u = 6 ( ∫ 1 u 2 + 1 d u + ∫ u 6 d u − ∫ u 4 d u + ∫ u 2 d u − ∫ 1 ⋅ d u ) = 6 tan − 1 u + 6 u 7 7 − 6 u 5 5 + 2 u 2 − 6 u = { substitute back u = x 6 } = 6 x 7 6 7 − 6 x 5 6 5 + 2 x − 6 x 6 + 6 tan − 1 ( x 6 ) + C \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} ∫ 1 + 3 x x d x = { substitute u = 6 x and d u = 6 x 6 5 d x } = 6 ∫ 1 + u 2 u 8 d u = 6 ∫ ( u 6 − u 4 + u 2 + u 2 + 1 1 − 1 ) d u = 6 ( ∫ u 2 + 1 1 d u + ∫ u 6 d u − ∫ u 4 d u + ∫ u 2 d u − ∫ 1 ⋅ d u ) = 6 tan − 1 u + 7 6 u 7 − 5 6 u 5 + 2 u 2 − 6 u = { substitute back u = 6 x } = 7 6 x 6 7 − 5 6 x 6 5 + 2 x − 6 6 x + 6 tan − 1 ( 6 x ) + C
Answer:
6 x 7 6 7 − 6 x 5 6 5 + 2 x − 6 x 6 + 6 tan − 1 ( x 6 ) + C \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 7 6 x 6 7 − 5 6 x 6 5 + 2 x − 6 6 x + 6 tan − 1 ( 6 x ) + C
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