What is the integral of (x^4+8sqrt(x))/(x^2) ?

The integral of (x^4+8sqrt(x))/(x^2) is (x^3)/3-(16)/(sqrt(x))+C

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integral of (x^4+8sqrt(x))/(x^2)

\int \frac{x ^{4} + 8 x}{x ^{2}} d x

\frac{x ^{3}}{3} - \frac{16}{x} + C

Solution steps

\int \frac{x ^{4} + 8 x}{x ^{2}} d x

Expand

\frac{x ^{4} + 8 x}{x ^{2}} : x^{2} + \frac{8}{x ^{\frac{3}{2}}}

= \int x^{2} + \frac{8}{x ^{\frac{3}{2}}} 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 x^{2} d x + \int \frac{8}{x ^{\frac{3}{2}}} d x

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

\int \frac{8}{x ^{\frac{3}{2}}} d x = - \frac{16}{x}

= \frac{x ^{3}}{3} - \frac{16}{x}

Add a constant to the solution

= \frac{x ^{3}}{3} - \frac{16}{x} + C

What is the integral of (x^4+8sqrt(x))/(x^2) ?
The integral of (x^4+8sqrt(x))/(x^2) is (x^3)/3-(16)/(sqrt(x))+C