What is the integral of (5x^2)/(sqrt(3x^3-8)) ?

The integral of (5x^2)/(sqrt(3x^3-8)) is 10/9 sqrt(3x^3-8)+C

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integral of (5x^2)/(sqrt(3x^3-8))

\int \frac{5 x ^{2}}{3 x ^{3} - 8} d x

\frac{10}{9} 3 x^{3} - 8 + C

Solution steps

\int \frac{5 x ^{2}}{3 x ^{3} - 8} d x

Take the constant out:

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

= 5 \cdot \int \frac{x ^{2}}{3 x ^{3} - 8} d x

Apply u-substitution

: \int \frac{u}{2 x 3 u ^{\frac{3}{2}} - 8} d u

= 5 \cdot \int \frac{1}{3 3 u - 8} d u

Take the constant out:

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

= 5 \cdot \frac{1}{3} \cdot \int \frac{1}{3 u - 8} d u

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: 2 v^{\frac{1}{2}}

= 5 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot 2 v^{\frac{1}{2}}

Substitute back

= 5 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot 2 (3 x^{3} - 8)^{\frac{1}{2}}

Simplify

5 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot 2 (3 x^{3} - 8)^{\frac{1}{2}} : \frac{10}{9} 3 x^{3} - 8

= \frac{10}{9} 3 x^{3} - 8

Add a constant to the solution

= \frac{10}{9} 3 x^{3} - 8 + C

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What is the integral of (5x^2)/(sqrt(3x^3-8)) ?
The integral of (5x^2)/(sqrt(3x^3-8)) is 10/9 sqrt(3x^3-8)+C