What is the integral of x^2\sqrt[3]{4x+5} ?

The integral of x^2\sqrt[3]{4x+5} is 1/64 (3/10 (4x+5)^{10/3}-30/7 (4x+5)^{7/3}+75/4 (4x+5)^{4/3})+C

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integral of x^2\sqrt[3]{4x+5}

\frac{1}{64} (\frac{3}{10} (4 x + 5)^{\frac{10}{3}} - \frac{30}{7} (4 x + 5)^{\frac{7}{3}} + \frac{75}{4} (4 x + 5)^{\frac{4}{3}}) + C

Solution steps

Apply u-substitution

Take the constant out:

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

Expand

Apply the Sum Rule:

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

\int u^{\frac{7}{3}} d u = \frac{3}{10} u^{\frac{10}{3}}

\int 10 u^{\frac{4}{3}} d u = \frac{30}{7} u^{\frac{7}{3}}

= \frac{1}{64} (\frac{3}{10} u^{\frac{10}{3}} - \frac{30}{7} u^{\frac{7}{3}} + \frac{75}{4} u^{\frac{4}{3}})

Substitute back

u = 4 x + 5

= \frac{1}{64} (\frac{3}{10} (4 x + 5)^{\frac{10}{3}} - \frac{30}{7} (4 x + 5)^{\frac{7}{3}} + \frac{75}{4} (4 x + 5)^{\frac{4}{3}})

Add a constant to the solution

= \frac{1}{64} (\frac{3}{10} (4 x + 5)^{\frac{10}{3}} - \frac{30}{7} (4 x + 5)^{\frac{7}{3}} + \frac{75}{4} (4 x + 5)^{\frac{4}{3}}) + C

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What is the integral of x^2\sqrt[3]{4x+5} ?
The integral of x^2\sqrt[3]{4x+5} is 1/64 (3/10 (4x+5)^{10/3}-30/7 (4x+5)^{7/3}+75/4 (4x+5)^{4/3})+C