What is the integral of (1/(t^2)+sqrt(t)) ?

The integral of (1/(t^2)+sqrt(t)) is -1/t+2/3 t^{3/2}+C

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integral of (1/(t^2)+sqrt(t))

\int (\frac{1}{t ^{2}} + t) d t

- \frac{1}{t} + \frac{2}{3} t^{\frac{3}{2}} + C

Solution steps

\int \frac{1}{t ^{2}} + 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 \frac{1}{t ^{2}} d t + \int t d t

\int \frac{1}{t ^{2}} d t = - \frac{1}{t}

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

= - \frac{1}{t} + \frac{2}{3} t^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{t} + \frac{2}{3} t^{\frac{3}{2}} + C

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What is the integral of (1/(t^2)+sqrt(t)) ?
The integral of (1/(t^2)+sqrt(t)) is -1/t+2/3 t^{3/2}+C