What is the integral of (2x+10)/(\sqrt[3]{x^2+10x)} ?

The integral of (2x+10)/(\sqrt[3]{x^2+10x)} is (3\sqrt[3]{4})/8 (4x^2+40x)^{2/3}+C

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integral of (2x+10)/(\sqrt[3]{x^2+10x)}

\int \frac{2 x + 10}{3 x ^{2} + 10 x} d x

\frac{3 3 4}{8} (4 x^{2} + 40 x)^{\frac{2}{3}} + C

Solution steps

\int \frac{2 x + 10}{3 x ^{2} + 10 x} d x

Apply u-substitution

: \int \frac{3 4 u}{2 3 u ^{2} - 100} d u

= \int \frac{3 4 u}{2 3 u ^{2} - 100} d u

Take the constant out:

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

= \frac{3 4}{2} \cdot \int \frac{u}{3 u ^{2} - 100} d u

Apply u-substitution

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

= \frac{3 4}{2} \cdot \int \frac{1}{2 3 v} d v

Take the constant out:

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

= \frac{3 4}{2} \cdot \frac{1}{2} \cdot \int \frac{1}{3 v} d v

Apply the Power Rule

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

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

Substitute back

= \frac{3 4}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} ((2 x + 10)^{2} - 100)^{\frac{2}{3}}

Simplify

\frac{3 4}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} ((2 x + 10)^{2} - 100)^{\frac{2}{3}} : \frac{3 3 4}{8} (4 x^{2} + 40 x)^{\frac{2}{3}}

= \frac{3 3 4}{8} (4 x^{2} + 40 x)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3 3 4}{8} (4 x^{2} + 40 x)^{\frac{2}{3}} + C

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What is the integral of (2x+10)/(\sqrt[3]{x^2+10x)} ?
The integral of (2x+10)/(\sqrt[3]{x^2+10x)} is (3\sqrt[3]{4})/8 (4x^2+40x)^{2/3}+C