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

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

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

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

\frac{4 u ^{3}}{3} - \frac{16}{u} + C

Solution steps

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

Expand

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

= \int 4 u^{2} + \frac{8}{u ^{\frac{3}{2}}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 4 u^{2} d u + \int \frac{8}{u ^{\frac{3}{2}}} d u

\int 4 u^{2} d u = \frac{4 u ^{3}}{3}

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

= \frac{4 u ^{3}}{3} - \frac{16}{u}

Add a constant to the solution

= \frac{4 u ^{3}}{3} - \frac{16}{u} + C

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