What is the integral of t^2-6sqrt(t) ?

The integral of t^2-6sqrt(t) is (t^3)/3-4t^{3/2}+C

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integral of t^2-6sqrt(t)

\int t^{2} - 6 t d t

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

Solution steps

\int t^{2} - 6 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^{2} d t - \int 6 t d t

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

\int 6 t d t = 4 t^{\frac{3}{2}}

= \frac{t ^{3}}{3} - 4 t^{\frac{3}{2}}

Add a constant to the solution

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

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What is the integral of t^2-6sqrt(t) ?
The integral of t^2-6sqrt(t) is (t^3)/3-4t^{3/2}+C