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

The integral of t^{(-1)/2}(t^2-t+2) is 2/5 t^{5/2}-2/3 t^{3/2}+4t^{1/2}+C

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integral of t^{(-1)/2}(t^2-t+2)

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

\frac{2}{5} t^{\frac{5}{2}} - \frac{2}{3} t^{\frac{3}{2}} + 4 t^{\frac{1}{2}} + C

Solution steps

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

Simplify

t^{\frac{- 1}{2}} (t^{2} - t + 2) : \frac{t ^{2} - t + 2}{t ^{\frac{1}{2}}}

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

Expand

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

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

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

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

\int \frac{2}{t ^{\frac{1}{2}}} d t = 4 t^{\frac{1}{2}}

= \frac{2}{5} t^{\frac{5}{2}} - \frac{2}{3} t^{\frac{3}{2}} + 4 t^{\frac{1}{2}}

Add a constant to the solution

= \frac{2}{5} t^{\frac{5}{2}} - \frac{2}{3} t^{\frac{3}{2}} + 4 t^{\frac{1}{2}} + C

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What is the integral of t^{(-1)/2}(t^2-t+2) ?
The integral of t^{(-1)/2}(t^2-t+2) is 2/5 t^{5/2}-2/3 t^{3/2}+4t^{1/2}+C