What is the integral of (4/(x^3)+6sqrt(x)) ?

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

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integral of (4/(x^3)+6sqrt(x))

\int (\frac{4}{x ^{3}} + 6 x) d x

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

Solution steps

\int \frac{4}{x ^{3}} + 6 x 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 \frac{4}{x ^{3}} d x + \int 6 x d x

\int \frac{4}{x ^{3}} d x = - \frac{2}{x ^{2}}

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

= - \frac{2}{x ^{2}} + 4 x^{\frac{3}{2}}

Add a constant to the solution

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

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