What is the integral of (t+2t^2)/(sqrt(t)) ?

The integral of (t+2t^2)/(sqrt(t)) is 2/3 t^{3/2}+4/5 t^{5/2}+C

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integral of (t+2t^2)/(sqrt(t))

\int \frac{t + 2 t ^{2}}{t} d t

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

Solution steps

\int \frac{t + 2 t ^{2}}{t} d t

Expand

\frac{t + 2 t ^{2}}{t} : \frac{t}{t} + \frac{2 t ^{2}}{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 \frac{t}{t} d t + \int \frac{2 t ^{2}}{t} d t

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

\int \frac{2 t ^{2}}{t} d t = \frac{4}{5} t^{\frac{5}{2}}

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

Add a constant to the solution

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

\frac{x d y}{d x} = 4 y

n = 1 \sum \infty \frac{1}{1 + ln ( n )}

\frac{\partial}{\partial x} (3 x^{4} y)

\frac{d}{d x} (\frac{4 x ^{4} - 7 x}{x ^{3} - 8})

f (x) = 2 x^{2} - 9 x + 10, at x = 3

What is the integral of (t+2t^2)/(sqrt(t)) ?
The integral of (t+2t^2)/(sqrt(t)) is 2/3 t^{3/2}+4/5 t^{5/2}+C