What is the integral of 4/(\sqrt[3]{7x+4)} ?

The integral of 4/(\sqrt[3]{7x+4)} is 6/7 (7x+4)^{2/3}+C

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integral of 4/(\sqrt[3]{7x+4)}

\frac{6}{7} (7 x + 4)^{\frac{2}{3}} + C

Solution steps

Take the constant out:

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

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

= 4 \cdot \frac{1}{7} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = 7 x + 4

= 4 \cdot \frac{1}{7} \cdot \frac{3}{2} (7 x + 4)^{\frac{2}{3}}

Simplify

4 \cdot \frac{1}{7} \cdot \frac{3}{2} (7 x + 4)^{\frac{2}{3}} : \frac{6}{7} (7 x + 4)^{\frac{2}{3}}

= \frac{6}{7} (7 x + 4)^{\frac{2}{3}}

Add a constant to the solution

= \frac{6}{7} (7 x + 4)^{\frac{2}{3}} + C

What is the integral of 4/(\sqrt[3]{7x+4)} ?
The integral of 4/(\sqrt[3]{7x+4)} is 6/7 (7x+4)^{2/3}+C