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

The integral of (\sqrt[3]{x}+1)^2 is 3/5 x^{5/3}+3/2 x^{4/3}+x+C

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

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

\frac{3}{5} x^{\frac{5}{3}} + \frac{3}{2} x^{\frac{4}{3}} + x + C

Solution steps

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

Expand

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

= \int x^{\frac{2}{3}} + 2 3 x + 1 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{2}{3}} d x + \int 2 3 x d x + \int 1 d x

\int x^{\frac{2}{3}} d x = \frac{3}{5} x^{\frac{5}{3}}

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

\int 1 d x = x

= \frac{3}{5} x^{\frac{5}{3}} + \frac{3}{2} x^{\frac{4}{3}} + x

Add a constant to the solution

= \frac{3}{5} x^{\frac{5}{3}} + \frac{3}{2} x^{\frac{4}{3}} + x + C

t^{2} y^{^{'}} + t y = 3, y (1) = 6

f (x) = 9 x

y = - x, y = \frac{1}{2} x - \frac{3}{2}, x = - 1, x = 3

\frac{x ^{2}}{( 3 + x )}

\int \frac{1}{9 x + 4} d x

What is the integral of (\sqrt[3]{x}+1)^2 ?
The integral of (\sqrt[3]{x}+1)^2 is 3/5 x^{5/3}+3/2 x^{4/3}+x+C