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)}

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

- 1.5 sin (1.5 x)

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\int sin (x) \cdot sin (2 \cdot x) d x

\frac{x - x}{x ^{\frac{1}{4}}}

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