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

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

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

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

Solution steps

Simplify

x^{2} : xx

Apply u-substitution

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{3}{10} u^{\frac{10}{3}} + \frac{30}{7} u^{\frac{7}{3}} + \frac{75}{4} u^{\frac{4}{3}}

Substitute back

u = x - 5

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

Add a constant to the solution

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

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