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

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

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integral of ((x+3))/(\sqrt[3]{x^2+6x)}

\int \frac{( x + 3 )}{3 x ^{2} + 6 x} d x

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

Solution steps

\int \frac{x + 3}{3 x ^{2} + 6 x} d x

Apply u-substitution

: \int \frac{1}{2 3 u} d u

= \int \frac{1}{2 3 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{3 u} d u

Apply the Power Rule

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

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

Substitute back

u = x^{2} + 6 x

= \frac{1}{2} \cdot \frac{3}{2} (x^{2} + 6 x)^{\frac{2}{3}}

Simplify

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

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

Add a constant to the solution

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

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