What is the integral of ((8q^3-3q^2-2q)/(4q)) ?

The integral of ((8q^3-3q^2-2q)/(4q)) is 1/4 ((8q^3)/3-(3q^2)/2-2q)+C

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integral of ((8q^3-3q^2-2q)/(4q))

\int (\frac{8 q ^{3} - 3 q ^{2} - 2 q}{4 q}) d q

\frac{1}{4} (\frac{8 q ^{3}}{3} - \frac{3 q ^{2}}{2} - 2 q) + C

Solution steps

\int \frac{8 q ^{3} - 3 q ^{2} - 2 q}{4 q} d q

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{8 q ^{3} - 3 q ^{2} - 2 q}{q} d q

Simplify

\frac{8 q ^{3} - 3 q ^{2} - 2 q}{q} : 8 q^{2} - 3 q - 2

= \frac{1}{4} \cdot \int 8 q^{2} - 3 q - 2 d q

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}{4} (\int 8 q^{2} d q - \int 3 q d q - \int 2 d q)

\int 8 q^{2} d q = \frac{8 q ^{3}}{3}

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

\int 2 d q = 2 q

= \frac{1}{4} (\frac{8 q ^{3}}{3} - \frac{3 q ^{2}}{2} - 2 q)

Add a constant to the solution

= \frac{1}{4} (\frac{8 q ^{3}}{3} - \frac{3 q ^{2}}{2} - 2 q) + C

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What is the integral of ((8q^3-3q^2-2q)/(4q)) ?
The integral of ((8q^3-3q^2-2q)/(4q)) is 1/4 ((8q^3)/3-(3q^2)/2-2q)+C