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

The integral of x^2\sqrt[3]{x+4} is 3/10 (x+4)^{10/3}-24/7 (x+4)^{7/3}+12(x+4)^{4/3}+C

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

\int x^{2} 3 x + 4 d x

\frac{3}{10} (x + 4)^{\frac{10}{3}} - \frac{24}{7} (x + 4)^{\frac{7}{3}} + 12 (x + 4)^{\frac{4}{3}} + C

Solution steps

\int x^{2} 3 x + 4 d x

Apply u-substitution

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

= \int (u - 4)^{2} 3 u d u

Expand

(u - 4)^{2} 3 u : u^{\frac{7}{3}} - 8 u^{\frac{4}{3}} + 16 3 u

= \int u^{\frac{7}{3}} - 8 u^{\frac{4}{3}} + 16 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

= \int u^{\frac{7}{3}} d u - \int 8 u^{\frac{4}{3}} d u + \int 16 3 u d u

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

\int 8 u^{\frac{4}{3}} d u = \frac{24}{7} u^{\frac{7}{3}}

\int 16 3 u d u = 12 u^{\frac{4}{3}}

= \frac{3}{10} u^{\frac{10}{3}} - \frac{24}{7} u^{\frac{7}{3}} + 12 u^{\frac{4}{3}}

Substitute back

u = x + 4

= \frac{3}{10} (x + 4)^{\frac{10}{3}} - \frac{24}{7} (x + 4)^{\frac{7}{3}} + 12 (x + 4)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{10} (x + 4)^{\frac{10}{3}} - \frac{24}{7} (x + 4)^{\frac{7}{3}} + 12 (x + 4)^{\frac{4}{3}} + C

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What is the integral of x^2\sqrt[3]{x+4} ?
The integral of x^2\sqrt[3]{x+4} is 3/10 (x+4)^{10/3}-24/7 (x+4)^{7/3}+12(x+4)^{4/3}+C