What is the integral of (1+8v)/(-16v^2) ?

The integral of (1+8v)/(-16v^2) is -1/16 (-1/v+8ln|v|)+C

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integral of (1+8v)/(-16v^2)

\int \frac{1 + 8 v}{- 16 v ^{2}} d v

- \frac{1}{16} (- \frac{1}{v} + 8 ln ∣ v ∣) + C

Solution steps

\int \frac{1 + 8 v}{- 16 v ^{2}} d v

Take the constant out:

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

= \frac{1}{- 16} \cdot \int \frac{1 + 8 v}{v ^{2}} d v

Simplify

= (- \frac{1}{16}) \cdot \int \frac{1 + 8 v}{v ^{2}} d v

Expand

\frac{1 + 8 v}{v ^{2}} : \frac{1}{v ^{2}} + \frac{8}{v}

= (- \frac{1}{16}) \cdot \int \frac{1}{v ^{2}} + \frac{8}{v} d v

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}{16}) (\int \frac{1}{v ^{2}} d v + \int \frac{8}{v} d v)

\int \frac{1}{v ^{2}} d v = - \frac{1}{v}

\int \frac{8}{v} d v = 8 ln ∣ v ∣

= (- \frac{1}{16}) (- \frac{1}{v} + 8 ln ∣ v ∣)

Simplify

= - \frac{1}{16} (- \frac{1}{v} + 8 ln ∣ v ∣)

Add a constant to the solution

= - \frac{1}{16} (- \frac{1}{v} + 8 ln ∣ v ∣) + C

What is the integral of (1+8v)/(-16v^2) ?
The integral of (1+8v)/(-16v^2) is -1/16 (-1/v+8ln|v|)+C