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

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

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

\frac{4}{9} (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

Apply the Power Rule

: \frac{4}{3} u^{\frac{3}{4}}

= \frac{1}{3} \cdot \frac{4}{3} u^{\frac{3}{4}}

Substitute back

u = 3 x + 1

= \frac{1}{3} \cdot \frac{4}{3} (3 x + 1)^{\frac{3}{4}}

Simplify

\frac{1}{3} \cdot \frac{4}{3} (3 x + 1)^{\frac{3}{4}} : \frac{4}{9} (3 x + 1)^{\frac{3}{4}}

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

Add a constant to the solution

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

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