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

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

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integral of x\sqrt[3]{x-4}

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

Solution steps

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{4}{3}} d u = \frac{3}{7} u^{\frac{7}{3}}

= \frac{3}{7} u^{\frac{7}{3}} + 3 u^{\frac{4}{3}}

Substitute back

u = x - 4

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

Add a constant to the solution

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

\int \frac{x ^{2}}{12} d x

f (x) = 8 6 x^{2} + 7

\int (\frac{x + 2}{x ^{2} + 4 x}) d x

\frac{d}{d x} (\frac{6 x ^{2} + 8 x + 2}{x})

\int \frac{7 x}{( x ^{2} + 2 ) ^{2}} d x

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