What is the integral of sqrt(t)(t^2-3t+5) ?

The integral of sqrt(t)(t^2-3t+5) is 2/7 t^{7/2}-6/5 t^{5/2}+10/3 t^{3/2}+C

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integral of sqrt(t)(t^2-3t+5)

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

\frac{2}{7} t^{\frac{7}{2}} - \frac{6}{5} t^{\frac{5}{2}} + \frac{10}{3} t^{\frac{3}{2}} + C

Solution steps

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

Expand

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

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

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

\int 3 t^{\frac{3}{2}} d t = \frac{6}{5} t^{\frac{5}{2}}

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

= \frac{2}{7} t^{\frac{7}{2}} - \frac{6}{5} t^{\frac{5}{2}} + \frac{10}{3} t^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} t^{\frac{7}{2}} - \frac{6}{5} t^{\frac{5}{2}} + \frac{10}{3} t^{\frac{3}{2}} + C

f (x) = \frac{3}{x + 1}, at x = 2

\int sin^{2} (θ) cos^{5} (θ) d θ

\frac{\partial}{\partial x} (e^{5 x y})

\frac{d}{d x} (sec^{2} (x) tan^{2} (x))

What is the integral of sqrt(t)(t^2-3t+5) ?
The integral of sqrt(t)(t^2-3t+5) is 2/7 t^{7/2}-6/5 t^{5/2}+10/3 t^{3/2}+C