What is the integral of (u^4+3sqrt(u))/(u^2) ?

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

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integral of (u^4+3sqrt(u))/(u^2)

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

\frac{u ^{3}}{3} - \frac{6}{u} + C

Solution steps

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

Expand

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

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

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

\int \frac{3}{u ^{\frac{3}{2}}} d u = - \frac{6}{u}

= \frac{u ^{3}}{3} - \frac{6}{u}

Add a constant to the solution

= \frac{u ^{3}}{3} - \frac{6}{u} + C

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