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

The integral of (t^2-1)/(sqrt(t)) is 2/5 t^{5/2}-2sqrt(t)+C

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

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

\frac{2}{5} t^{\frac{5}{2}} - 2 t + C

Solution steps

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

Expand

\frac{t ^{2} - 1}{t} : t^{\frac{3}{2}} - \frac{1}{t}

= \int t^{\frac{3}{2}} - \frac{1}{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{3}{2}} d t - \int \frac{1}{t} d t

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

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

= \frac{2}{5} t^{\frac{5}{2}} - 2 t

Add a constant to the solution

= \frac{2}{5} t^{\frac{5}{2}} - 2 t + C

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What is the integral of (t^2-1)/(sqrt(t)) ?
The integral of (t^2-1)/(sqrt(t)) is 2/5 t^{5/2}-2sqrt(t)+C