What is the integral of \sqrt[3]{t}(5t^2-3t+2) ?

The integral of \sqrt[3]{t}(5t^2-3t+2) is 3/2 t^{10/3}-9/7 t^{7/3}+3/2 t^{4/3}+C

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integral of \sqrt[3]{t}(5t^2-3t+2)

\int 3 t (5 t^{2} - 3 t + 2) d t

\frac{3}{2} t^{\frac{10}{3}} - \frac{9}{7} t^{\frac{7}{3}} + \frac{3}{2} t^{\frac{4}{3}} + C

Solution steps

\int 3 t (5 t^{2} - 3 t + 2) d t

Expand

3 t (5 t^{2} - 3 t + 2) : 5 t^{\frac{7}{3}} - 3 t^{\frac{4}{3}} + 2 3 t

= \int 5 t^{\frac{7}{3}} - 3 t^{\frac{4}{3}} + 2 3 t d t

Apply the Sum Rule:

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

= \int 5 t^{\frac{7}{3}} d t - \int 3 t^{\frac{4}{3}} d t + \int 2 3 t d t

\int 5 t^{\frac{7}{3}} d t = \frac{3}{2} t^{\frac{10}{3}}

\int 3 t^{\frac{4}{3}} d t = \frac{9}{7} t^{\frac{7}{3}}

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

= \frac{3}{2} t^{\frac{10}{3}} - \frac{9}{7} t^{\frac{7}{3}} + \frac{3}{2} t^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{2} t^{\frac{10}{3}} - \frac{9}{7} t^{\frac{7}{3}} + \frac{3}{2} t^{\frac{4}{3}} + C

n = 1 \sum \infty 3 n + 1

\int_{0}^{1} 1 + (2 x)^{2} d x

y = 3 + 5 x

\frac{\partial}{\partial x} (cosh (x))

\int \frac{5 r ^{2}}{r + 1} d r

What is the integral of \sqrt[3]{t}(5t^2-3t+2) ?
The integral of \sqrt[3]{t}(5t^2-3t+2) is 3/2 t^{10/3}-9/7 t^{7/3}+3/2 t^{4/3}+C