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

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

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

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

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

Solution steps

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

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 x^{2} d x + \int \frac{1}{x ^{3}} d x

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

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

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

Add a constant to the solution

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

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