What is the integral of (9t^2+1/(sqrt(t^3))) ?

The integral of (9t^2+1/(sqrt(t^3))) is 3t^3-2/(sqrt(t))+C

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integral of (9t^2+1/(sqrt(t^3)))

\int (9 t^{2} + \frac{1}{t ^{3}}) d t

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

Solution steps

\int 9 t^{2} + \frac{1}{t ^{3}} 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 9 t^{2} d t + \int \frac{1}{t ^{3}} d t

\int 9 t^{2} d t = 3 t^{3}

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

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

Add a constant to the solution

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

What is the integral of (9t^2+1/(sqrt(t^3))) ?
The integral of (9t^2+1/(sqrt(t^3))) is 3t^3-2/(sqrt(t))+C