What is the integral of 1/(15+2x-x^2) ?

The integral of 1/(15+2x-x^2) is 1/8 ln|x+3|-1/8 ln|x-5|+C

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integral of 1/(15+2x-x^2)

\int \frac{1}{15 + 2 x - x ^{2}} d x

\frac{1}{8} ln ∣ x + 3 ∣ - \frac{1}{8} ln ∣ x - 5 ∣ + C

Solution steps

\int \frac{1}{15 + 2 x - x ^{2}} d x

Take the partial fraction of

\frac{1}{15 + 2 x - x ^{2}} : \frac{1}{8 ( x + 3 )} - \frac{1}{8 ( x - 5 )}

= \int \frac{1}{8 ( x + 3 )} - \frac{1}{8 ( x - 5 )} d x

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}{8 ( x + 3 )} d x - \int \frac{1}{8 ( x - 5 )} d x

\int \frac{1}{8 ( x + 3 )} d x = \frac{1}{8} ln ∣ x + 3 ∣

\int \frac{1}{8 ( x - 5 )} d x = \frac{1}{8} ln ∣ x - 5 ∣

= \frac{1}{8} ln ∣ x + 3 ∣ - \frac{1}{8} ln ∣ x - 5 ∣

Add a constant to the solution

= \frac{1}{8} ln ∣ x + 3 ∣ - \frac{1}{8} ln ∣ x - 5 ∣ + C

What is the integral of 1/(15+2x-x^2) ?
The integral of 1/(15+2x-x^2) is 1/8 ln|x+3|-1/8 ln|x-5|+C