What is the integral of (6x+5)/(\sqrt[3]{3x^2+5x)} ?

The integral of (6x+5)/(\sqrt[3]{3x^2+5x)} is (\sqrt[3]{3})/(4\sqrt[3]{2)}(36x^2+60x)^{2/3}+C

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integral of (6x+5)/(\sqrt[3]{3x^2+5x)}

\int \frac{6 x + 5}{3 3 x ^{2} + 5 x} d x

\frac{3 3}{4 3 2} (36 x^{2} + 60 x)^{\frac{2}{3}} + C

Solution steps

\int \frac{6 x + 5}{3 3 x ^{2} + 5 x} d x

Apply u-substitution

: \int \frac{u}{2 ^{\frac{1}{3}} \cdot 3 ^{\frac{2}{3}} 3 u ^{2} - 25} d u

= \int \frac{u}{2 ^{\frac{1}{3}} \cdot 3 ^{\frac{2}{3}} 3 u ^{2} - 25} d u

Take the constant out:

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

= \frac{1}{2 ^{\frac{1}{3}} \cdot 3 ^{\frac{2}{3}}} \cdot \int \frac{u}{3 u ^{2} - 25} d u

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

= \frac{1}{2 ^{\frac{1}{3}} \cdot 3 ^{\frac{2}{3}}} \cdot \frac{1}{2} \cdot \frac{3}{2} ((6 x + 5)^{2} - 25)^{\frac{2}{3}}

Simplify

\frac{1}{2 ^{\frac{1}{3}} \cdot 3 ^{\frac{2}{3}}} \cdot \frac{1}{2} \cdot \frac{3}{2} ((6 x + 5)^{2} - 25)^{\frac{2}{3}} : \frac{3 3}{4 3 2} (36 x^{2} + 60 x)^{\frac{2}{3}}

= \frac{3 3}{4 3 2} (36 x^{2} + 60 x)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3 3}{4 3 2} (36 x^{2} + 60 x)^{\frac{2}{3}} + C

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What is the integral of (6x+5)/(\sqrt[3]{3x^2+5x)} ?
The integral of (6x+5)/(\sqrt[3]{3x^2+5x)} is (\sqrt[3]{3})/(4\sqrt[3]{2)}(36x^2+60x)^{2/3}+C