What is the integral of \sqrt[3]{x}*(x+1)^2 ?

The integral of \sqrt[3]{x}*(x+1)^2 is 3/10 x^{10/3}+6/7 x^{7/3}+3/4 x^{4/3}+C

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integral of \sqrt[3]{x}*(x+1)^2

\int 3 x \cdot (x + 1)^{2} d x

\frac{3}{10} x^{\frac{10}{3}} + \frac{6}{7} x^{\frac{7}{3}} + \frac{3}{4} x^{\frac{4}{3}} + C

Solution steps

\int 3 x (x + 1)^{2} d x

Expand

3 x (x + 1)^{2} : x^{\frac{7}{3}} + 2 x^{\frac{4}{3}} + 3 x

= \int x^{\frac{7}{3}} + 2 x^{\frac{4}{3}} + 3 x d x

Apply the Sum Rule:

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

= \int x^{\frac{7}{3}} d x + \int 2 x^{\frac{4}{3}} d x + \int 3 x d x

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

\int 2 x^{\frac{4}{3}} d x = \frac{6}{7} x^{\frac{7}{3}}

\int 3 x d x = \frac{3}{4} x^{\frac{4}{3}}

= \frac{3}{10} x^{\frac{10}{3}} + \frac{6}{7} x^{\frac{7}{3}} + \frac{3}{4} x^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{10} x^{\frac{10}{3}} + \frac{6}{7} x^{\frac{7}{3}} + \frac{3}{4} x^{\frac{4}{3}} + C

(1 + y^{2} + x y^{2}) d x + (x^{2} y + y + 2 x y) d y = 0

\frac{d}{d x} (x^{0.6})

a re a y = x, y = 12 x, x = 9

l a pl a ce t r an s f or m 5 x

t an g e n t f (x) = \frac{6}{x - 1}

What is the integral of \sqrt[3]{x}*(x+1)^2 ?
The integral of \sqrt[3]{x}*(x+1)^2 is 3/10 x^{10/3}+6/7 x^{7/3}+3/4 x^{4/3}+C