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

The integral of ((u^3+2u^2-u)/(3u)) is 1/3 ((u^3)/3+u^2-u)+C

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integral of ((u^3+2u^2-u)/(3u))

\int (\frac{u ^{3} + 2 u ^{2} - u}{3 u}) d u

\frac{1}{3} (\frac{u ^{3}}{3} + u^{2} - u) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{u ^{3} + 2 u ^{2} - u}{u} d u

Expand

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

Apply the Sum Rule:

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

= \frac{1}{3} (\int \frac{u ^{3}}{u} d u + \int \frac{2 u ^{2}}{u} d u - \int \frac{u}{u} d u)

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

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

\int \frac{u}{u} d u = u

= \frac{1}{3} (\frac{u ^{3}}{3} + u^{2} - u)

Add a constant to the solution

= \frac{1}{3} (\frac{u ^{3}}{3} + u^{2} - u) + C

\int \frac{3 x ^{2}}{( 196 + x ^{2} ) ^{2}} d x

\int a x^{4} + b x^{3} + 3 c d x

\int \frac{x}{x ^{4} - 2 x ^{3} + 2 x ^{2} - 2 x + 1} d x

\int (5^{y} + \frac{6 x}{y + 4}) d x

\int 10 x^{- 1} d x

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