What is the integral of (x^2+x)\sqrt[3]{x+7} ?

The integral of (x^2+x)\sqrt[3]{x+7} is 3/10 (x+7)^{10/3}-39/7 (x+7)^{7/3}+63/2 (x+7)^{4/3}+C

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integral of (x^2+x)\sqrt[3]{x+7}

\frac{3}{10} (x + 7)^{\frac{10}{3}} - \frac{39}{7} (x + 7)^{\frac{7}{3}} + \frac{63}{2} (x + 7)^{\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{7}{3}} d u = \frac{3}{10} u^{\frac{10}{3}}

\int 13 u^{\frac{4}{3}} d u = \frac{39}{7} u^{\frac{7}{3}}

= \frac{3}{10} u^{\frac{10}{3}} - \frac{39}{7} u^{\frac{7}{3}} + \frac{63}{2} u^{\frac{4}{3}}

Substitute back

u = x + 7

= \frac{3}{10} (x + 7)^{\frac{10}{3}} - \frac{39}{7} (x + 7)^{\frac{7}{3}} + \frac{63}{2} (x + 7)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{10} (x + 7)^{\frac{10}{3}} - \frac{39}{7} (x + 7)^{\frac{7}{3}} + \frac{63}{2} (x + 7)^{\frac{4}{3}} + C

y = \frac{x}{x}

t^{2} y^{^{''}} + 17 t y^{^{'}} + 64 y = 0

x \to 0 lim (x (0))

\frac{\partial}{\partial y} (sin (- 2 x) cos (6 y))

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the integral of (x^2+x)\sqrt[3]{x+7} ?
The integral of (x^2+x)\sqrt[3]{x+7} is 3/10 (x+7)^{10/3}-39/7 (x+7)^{7/3}+63/2 (x+7)^{4/3}+C