What is the integral of 18x^2(6x^3+5)^{1/2} ?

The integral of 18x^2(6x^3+5)^{1/2} is 2/3 (6x^3+5)^{3/2}+C

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integral of 18x^2(6x^3+5)^{1/2}

\int 18 x^{2} (6 x^{3} + 5)^{\frac{1}{2}} d x

\frac{2}{3} (6 x^{3} + 5)^{\frac{3}{2}} + C

Solution steps

\int 18 x^{2} (6 x^{3} + 5)^{\frac{1}{2}} 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} (6 x^{3} + 5)^{\frac{1}{2}} d x

Apply u-substitution

: \int \frac{u ( 6 u ^{\frac{3}{2}} + 5 ) ^{\frac{1}{2}}}{2 x} d u

= 18 \cdot \int \frac{( 6 u + 5 ) ^{\frac{1}{2}}}{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 (6 u + 5)^{\frac{1}{2}} d u

Apply u-substitution

: \int \frac{v ^{\frac{1}{2}}}{6} d v

= 18 \cdot \frac{1}{3} \cdot \int \frac{v ^{\frac{1}{2}}}{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 v^{\frac{1}{2}} d v

Apply the Power Rule

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

= 18 \cdot \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{2}{3} v^{\frac{3}{2}}

Substitute back

= 18 \cdot \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{2}{3} (6 x^{3} + 5)^{\frac{3}{2}}

Simplify

18 \cdot \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{2}{3} (6 x^{3} + 5)^{\frac{3}{2}} : \frac{2}{3} (6 x^{3} + 5)^{\frac{3}{2}}

= \frac{2}{3} (6 x^{3} + 5)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (6 x^{3} + 5)^{\frac{3}{2}} + C

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What is the integral of 18x^2(6x^3+5)^{1/2} ?
The integral of 18x^2(6x^3+5)^{1/2} is 2/3 (6x^3+5)^{3/2}+C