What is the integral of 1/(v^2+8v-9) ?

The integral of 1/(v^2+8v-9) is 1/10 ln|v-1|-1/10 ln|v+9|+C

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integral of 1/(v^2+8v-9)

\int \frac{1}{v ^{2} + 8 v - 9} d v

\frac{1}{10} ln ∣ v - 1 ∣ - \frac{1}{10} ln ∣ v + 9 ∣ + C

Solution steps

\int \frac{1}{v ^{2} + 8 v - 9} d v

Take the partial fraction of

\frac{1}{v ^{2} + 8 v - 9} : \frac{1}{10 ( v - 1 )} - \frac{1}{10 ( v + 9 )}

= \int \frac{1}{10 ( v - 1 )} - \frac{1}{10 ( v + 9 )} 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

= \int \frac{1}{10 ( v - 1 )} d v - \int \frac{1}{10 ( v + 9 )} d v

\int \frac{1}{10 ( v - 1 )} d v = \frac{1}{10} ln ∣ v - 1 ∣

\int \frac{1}{10 ( v + 9 )} d v = \frac{1}{10} ln ∣ v + 9 ∣

= \frac{1}{10} ln ∣ v - 1 ∣ - \frac{1}{10} ln ∣ v + 9 ∣

Add a constant to the solution

= \frac{1}{10} ln ∣ v - 1 ∣ - \frac{1}{10} ln ∣ v + 9 ∣ + C

\int - \frac{2}{x + 1} d x

\int x (3 x - 2) d x

\frac{d y}{d x} + x y = y

\int_{- 1}^{9} 9 - x^{2} + 8 x d x

y^{^{'}} = \frac{5 x sec ( \frac{y}{x} ) + y}{x}

What is the integral of 1/(v^2+8v-9) ?
The integral of 1/(v^2+8v-9) is 1/10 ln|v-1|-1/10 ln|v+9|+C