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

The integral of 0.06q-0.6q-(7700)/(q^2) is -0.27q^2+(7700)/q+C

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

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

- 0.27 q^{2} + \frac{7700}{q} + C

Solution steps

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

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

\int 0.6 q d q = 0.3 q^{2}

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

= 0.03 q^{2} - 0.3 q^{2} - (- \frac{7700}{q})

Simplify

= - 0.27 q^{2} + \frac{7700}{q}

Add a constant to the solution

= - 0.27 q^{2} + \frac{7700}{q} + C

\frac{d y}{d x} = a - b y^{2}

\int_{1}^{2} 4 x^{2} - 3 d x

x \to 2 - lim (k x^{2})

x \to 0 lim (\frac{- 1}{5 x + 25})

\frac{\partial}{\partial x} (2 x arctan (\frac{y}{z}))

What is the integral of 0.06q-0.6q-(7700)/(q^2) ?
The integral of 0.06q-0.6q-(7700)/(q^2) is -0.27q^2+(7700)/q+C