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

The integral of 1/(\sqrt[3]{4x+7)} is 3/8 (4x+7)^{2/3}+C

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integral of 1/(\sqrt[3]{4x+7)}

\int \frac{1}{3 4 x + 7} d x

\frac{3}{8} (4 x + 7)^{\frac{2}{3}} + C

Solution steps

\int \frac{1}{3 4 x + 7} d x

Apply u-substitution

: \int \frac{1}{4 3 u} d u

= \int \frac{1}{4 3 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{3 u} d u

Apply the Power Rule

: \frac{3}{2} u^{\frac{2}{3}}

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

Substitute back

u = 4 x + 7

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

Simplify

\frac{1}{4} \cdot \frac{3}{2} (4 x + 7)^{\frac{2}{3}} : \frac{3}{8} (4 x + 7)^{\frac{2}{3}}

= \frac{3}{8} (4 x + 7)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{8} (4 x + 7)^{\frac{2}{3}} + C

2 y^{^{''}} + 13 y^{^{'}} - 7 y = 0

\int \frac{1}{x ^{2} + t ^{2}} d x

\frac{d}{d x} (\frac{1}{3} tan^{3} (x) - tan (x) + x)

f (x) = (x + 9) (x^{2} - 9)

\frac{6}{s ^{4} ( s - 2 ) ^{2}}

What is the integral of 1/(\sqrt[3]{4x+7)} ?
The integral of 1/(\sqrt[3]{4x+7)} is 3/8 (4x+7)^{2/3}+C