What is the integral of (x+1)\sqrt[3]{4-x} ?

The integral of (x+1)\sqrt[3]{4-x} is 3/7 (4-x)^{7/3}-15/4 (4-x)^{4/3}+C

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integral of (x+1)\sqrt[3]{4-x}

\frac{3}{7} (4 - x)^{\frac{7}{3}} - \frac{15}{4} (4 - x)^{\frac{4}{3}} + C

Solution steps

Expand

Apply the Sum Rule:

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

= \frac{3}{7} (4 - x)^{\frac{7}{3}} - 3 (4 - x)^{\frac{4}{3}} - \frac{3}{4} (4 - x)^{\frac{4}{3}}

Add similar elements:

- 3 (- x + 4)^{\frac{4}{3}} - \frac{3}{4} (- x + 4)^{\frac{4}{3}} = - \frac{15}{4} (- x + 4)^{\frac{4}{3}}

= \frac{3}{7} (4 - x)^{\frac{7}{3}} - \frac{15}{4} (4 - x)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{7} (4 - x)^{\frac{7}{3}} - \frac{15}{4} (4 - x)^{\frac{4}{3}} + C

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What is the integral of (x+1)\sqrt[3]{4-x} ?
The integral of (x+1)\sqrt[3]{4-x} is 3/7 (4-x)^{7/3}-15/4 (4-x)^{4/3}+C