What is the integral of x/(\sqrt[4]{x+8)} ?

The integral of x/(\sqrt[4]{x+8)} is 4/7 (x+8)^{7/4}-32/3 (x+8)^{3/4}+C

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integral of x/(\sqrt[4]{x+8)}

\frac{4}{7} (x + 8)^{\frac{7}{4}} - \frac{32}{3} (x + 8)^{\frac{3}{4}} + C

Solution steps

Apply u-substitution

Expand

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

\int u^{\frac{3}{4}} d u = \frac{4}{7} u^{\frac{7}{4}}

= \frac{4}{7} u^{\frac{7}{4}} - \frac{32}{3} u^{\frac{3}{4}}

Substitute back

u = x + 8

= \frac{4}{7} (x + 8)^{\frac{7}{4}} - \frac{32}{3} (x + 8)^{\frac{3}{4}}

Add a constant to the solution

= \frac{4}{7} (x + 8)^{\frac{7}{4}} - \frac{32}{3} (x + 8)^{\frac{3}{4}} + C

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What is the integral of x/(\sqrt[4]{x+8)} ?
The integral of x/(\sqrt[4]{x+8)} is 4/7 (x+8)^{7/4}-32/3 (x+8)^{3/4}+C