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^{^{'}} = e^{3 x} - y

\int 3 tan^{3} (x) d x

n = 0 \sum \infty \frac{7 ^{n}}{4}

x^{^{''}} + 2 x^{^{'}} + 2 x = 0

\frac{d ^{2}}{d x ^{2}} (- sin (3 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