What is the integral of (18x^2\sqrt[5]{6x^3+5}) ?

The integral of (18x^2\sqrt[5]{6x^3+5}) is 5/6 (6x^3+5)^{6/5}+C

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integral of (18x^2\sqrt[5]{6x^3+5})

\int (18 x^{2} 5 6 x^{3} + 5) d x

\frac{5}{6} (6 x^{3} + 5)^{\frac{6}{5}} + C

Solution steps

\int 18 x^{2} 5 6 x^{3} + 5 d x

Take the constant out:

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

= 18 \cdot \int x^{2} 5 6 x^{3} + 5 d x

Apply u-substitution

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

= 18 \cdot \int \frac{5 6 u + 5}{3} d u

Take the constant out:

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

= 18 \cdot \frac{1}{3} \cdot \int 5 6 u + 5 d u

Apply u-substitution

: \int \frac{5 v}{6} d v

= 18 \cdot \frac{1}{3} \cdot \int \frac{5 v}{6} d v

Take the constant out:

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

= 18 \cdot \frac{1}{3} \cdot \frac{1}{6} \cdot \int 5 v d v

Apply the Power Rule

: \frac{5}{6} v^{\frac{6}{5}}

= 18 \cdot \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{5}{6} v^{\frac{6}{5}}

Substitute back

= 18 \cdot \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{5}{6} (6 x^{3} + 5)^{\frac{6}{5}}

Simplify

18 \cdot \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{5}{6} (6 x^{3} + 5)^{\frac{6}{5}} : \frac{5}{6} (6 x^{3} + 5)^{\frac{6}{5}}

= \frac{5}{6} (6 x^{3} + 5)^{\frac{6}{5}}

Add a constant to the solution

= \frac{5}{6} (6 x^{3} + 5)^{\frac{6}{5}} + C

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What is the integral of (18x^2\sqrt[5]{6x^3+5}) ?
The integral of (18x^2\sqrt[5]{6x^3+5}) is 5/6 (6x^3+5)^{6/5}+C