What is the integral of 0.06q-0.6-(7700)/(q^2) ?

The integral of 0.06q-0.6-(7700)/(q^2) is 0.03q^2-0.6q+(7700)/q+C

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integral of 0.06q-0.6-(7700)/(q^2)

\int 0.06 q - 0.6 - \frac{7700}{q ^{2}} d q

0.03 q^{2} - 0.6 q + \frac{7700}{q} + C

Solution steps

\int 0.06 q - 0.6 - \frac{7700}{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

= \int 0.06 q d q - \int 0.6 d q - \int \frac{7700}{q ^{2}} d q

\int 0.06 q d q = 0.03 q^{2}

\int 0.6 d q = 0.6 q

\int \frac{7700}{q ^{2}} d q = - \frac{7700}{q}

= 0.03 q^{2} - 0.6 q - (- \frac{7700}{q})

Simplify

= 0.03 q^{2} - 0.6 q + \frac{7700}{q}

Add a constant to the solution

= 0.03 q^{2} - 0.6 q + \frac{7700}{q} + C

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What is the integral of 0.06q-0.6-(7700)/(q^2) ?
The integral of 0.06q-0.6-(7700)/(q^2) is 0.03q^2-0.6q+(7700)/q+C