What is the integral of x/(\sqrt[4]{3x+1)} ?

The integral of x/(\sqrt[4]{3x+1)} is 1/9 (4/7 (3x+1)^{7/4}-4/3 (3x+1)^{3/4})+C

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integral of x/(\sqrt[4]{3x+1)}

\frac{1}{9} (\frac{4}{7} (3 x + 1)^{\frac{7}{4}} - \frac{4}{3} (3 x + 1)^{\frac{3}{4}}) + C

Solution steps

Apply u-substitution

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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{1}{9} (\frac{4}{7} u^{\frac{7}{4}} - \frac{4}{3} u^{\frac{3}{4}})

Substitute back

u = 3 x + 1

= \frac{1}{9} (\frac{4}{7} (3 x + 1)^{\frac{7}{4}} - \frac{4}{3} (3 x + 1)^{\frac{3}{4}})

Add a constant to the solution

= \frac{1}{9} (\frac{4}{7} (3 x + 1)^{\frac{7}{4}} - \frac{4}{3} (3 x + 1)^{\frac{3}{4}}) + C

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What is the integral of x/(\sqrt[4]{3x+1)} ?
The integral of x/(\sqrt[4]{3x+1)} is 1/9 (4/7 (3x+1)^{7/4}-4/3 (3x+1)^{3/4})+C