What is the integral of (3u-4)/5 ?

The integral of (3u-4)/5 is 1/5 ((3u^2)/2-4u)+C

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integral of (3u-4)/5

\int \frac{3 u - 4}{5} d u

\frac{1}{5} (\frac{3 u ^{2}}{2} - 4 u) + C

Solution steps

\int \frac{3 u - 4}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int 3 u - 4 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

= \frac{1}{5} (\int 3 u d u - \int 4 d u)

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

\int 4 d u = 4 u

= \frac{1}{5} (\frac{3 u ^{2}}{2} - 4 u)

Add a constant to the solution

= \frac{1}{5} (\frac{3 u ^{2}}{2} - 4 u) + C

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What is the integral of (3u-4)/5 ?
The integral of (3u-4)/5 is 1/5 ((3u^2)/2-4u)+C