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

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

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

\int \frac{1}{1 + 3 x + 1} d x

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

Solution steps

\int \frac{1}{1 + 3 x + 1} d x

Apply u-substitution

= \int \frac{1}{4 u} d u

Apply the Power Rule

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

Substitute back

u = x + 1

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

Add a constant to the solution

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

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