What is the integral of 5x[(9-4x^2)^2]^{1/3} ?

The integral of 5x[(9-4x^2)^2]^{1/3} is -3/8 (9-4x^2)^{5/3}+C

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integral of 5x[(9-4x^2)^2]^{1/3}

\int 5 x [(9 - 4 x^{2})^{2}]^{\frac{1}{3}} d x

- \frac{3}{8} (9 - 4 x^{2})^{\frac{5}{3}} + C

Solution steps

\int 5 x [(9 - 4 x^{2})^{2}]^{\frac{1}{3}} d x

Take the constant out:

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

= 5 \cdot \int x [(9 - 4 x^{2})^{2}]^{\frac{1}{3}} d x

Apply u-substitution

: \int \frac{( 9 - 4 u ) ^{\frac{2}{3}}}{2} d u

= 5 \cdot \int \frac{( 9 - 4 u ) ^{\frac{2}{3}}}{2} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int (9 - 4 u)^{\frac{2}{3}} d u

Apply u-substitution

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

= 5 \cdot \frac{1}{2} \cdot \int - \frac{v ^{\frac{2}{3}}}{4} d v

Take the constant out:

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

= 5 \cdot \frac{1}{2} (- \frac{1}{4} \cdot \int v^{\frac{2}{3}} d v)

Apply the Power Rule

: \frac{3}{5} v^{\frac{5}{3}}

= 5 \cdot \frac{1}{2} (- \frac{1}{4} \cdot \frac{3}{5} v^{\frac{5}{3}})

Substitute back

= 5 \cdot \frac{1}{2} (- \frac{1}{4} \cdot \frac{3}{5} (9 - 4 x^{2})^{\frac{5}{3}})

Simplify

5 \cdot \frac{1}{2} (- \frac{1}{4} \cdot \frac{3}{5} (9 - 4 x^{2})^{\frac{5}{3}}) : - \frac{3}{8} (9 - 4 x^{2})^{\frac{5}{3}}

= - \frac{3}{8} (9 - 4 x^{2})^{\frac{5}{3}}

Add a constant to the solution

= - \frac{3}{8} (9 - 4 x^{2})^{\frac{5}{3}} + C

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What is the integral of 5x[(9-4x^2)^2]^{1/3} ?
The integral of 5x[(9-4x^2)^2]^{1/3} is -3/8 (9-4x^2)^{5/3}+C