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

The integral of (4x^2)/(\sqrt[3]{5-5x)} is -3/(10\sqrt[3]{5)}(5x^2+6x+9)(-x+1)^{2/3}+C

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

\int \frac{4 x ^{2}}{3 5 - 5 x} d x

- \frac{3}{10 3 5} (5 x^{2} + 6 x + 9) (- x + 1)^{\frac{2}{3}} + C

Solution steps

\int \frac{4 x ^{2}}{3 5 - 5 x} d x

Take the constant out:

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

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

Apply u-substitution

: \int - \frac{( u - 5 ) ^{2}}{125 3 u} d u

= 4 \cdot \int - \frac{( u - 5 ) ^{2}}{125 3 u} d u

Take the constant out:

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

= 4 (- \frac{1}{125} \cdot \int \frac{( u - 5 ) ^{2}}{3 u} d u)

Expand

\frac{( u - 5 ) ^{2}}{3 u} : u^{\frac{5}{3}} - 10 u^{\frac{2}{3}} + \frac{25}{3 u}

= 4 (- \frac{1}{125} \cdot \int u^{\frac{5}{3}} - 10 u^{\frac{2}{3}} + \frac{25}{3 u} d u)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (- \frac{1}{125} (\int u^{\frac{5}{3}} d u - \int 10 u^{\frac{2}{3}} d u + \int \frac{25}{3 u} d u))

\int u^{\frac{5}{3}} d u = \frac{3}{8} u^{\frac{8}{3}}

\int 10 u^{\frac{2}{3}} d u = 6 u^{\frac{5}{3}}

\int \frac{25}{3 u} d u = \frac{75}{2} u^{\frac{2}{3}}

= 4 (- \frac{1}{125} (\frac{3}{8} u^{\frac{8}{3}} - 6 u^{\frac{5}{3}} + \frac{75}{2} u^{\frac{2}{3}}))

Substitute back

u = 5 - 5 x

= 4 (- \frac{1}{125} (\frac{3}{8} (5 - 5 x)^{\frac{8}{3}} - 6 (5 - 5 x)^{\frac{5}{3}} + \frac{75}{2} (5 - 5 x)^{\frac{2}{3}}))

Simplify

4 (- \frac{1}{125} (\frac{3}{8} (5 - 5 x)^{\frac{8}{3}} - 6 (5 - 5 x)^{\frac{5}{3}} + \frac{75}{2} (5 - 5 x)^{\frac{2}{3}})) : - \frac{3}{10 3 5} (5 x^{2} + 6 x + 9) (- x + 1)^{\frac{2}{3}}

= - \frac{3}{10 3 5} (5 x^{2} + 6 x + 9) (- x + 1)^{\frac{2}{3}}

Add a constant to the solution

= - \frac{3}{10 3 5} (5 x^{2} + 6 x + 9) (- x + 1)^{\frac{2}{3}} + C

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What is the integral of (4x^2)/(\sqrt[3]{5-5x)} ?
The integral of (4x^2)/(\sqrt[3]{5-5x)} is -3/(10\sqrt[3]{5)}(5x^2+6x+9)(-x+1)^{2/3}+C