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

y^{^{''}} + y^{^{'}} - 2 y = 6 e^{x}

4 y^{^{''}} + 8 y^{^{'}} + 3 y = 0

\frac{5 s}{( s ^{2} + 4 ) \cdot ( s + 7 )}

f (x) = (\frac{5 x ^{2} + 6 x + 1}{3 x - 5})^{\frac{3}{2}}

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

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