What is the integral of 1/(x^2-8x+15) ?

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

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integral of 1/(x^2-8x+15)

\int \frac{1}{x ^{2} - 8 x + 15} d x

- \frac{1}{2} ln ∣ x - 3 ∣ + \frac{1}{2} ln ∣ x - 5 ∣ + C

Solution steps

\int \frac{1}{x ^{2} - 8 x + 15} d x

Take the partial fraction of

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

= \int - \frac{1}{2 ( x - 3 )} + \frac{1}{2 ( 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}{2 ( x - 3 )} d x + \int \frac{1}{2 ( x - 5 )} d x

\int \frac{1}{2 ( x - 3 )} d x = \frac{1}{2} ln ∣ x - 3 ∣

\int \frac{1}{2 ( x - 5 )} d x = \frac{1}{2} ln ∣ x - 5 ∣

= - \frac{1}{2} ln ∣ x - 3 ∣ + \frac{1}{2} ln ∣ x - 5 ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ x - 3 ∣ + \frac{1}{2} ln ∣ x - 5 ∣ + C

(x y + 4 y^{2} + 2 x^{2}) d x - (x^{2}) d y = 0

y = e^{9}

\int + e^{- 3 x} sin (b x) d x

\frac{d}{d x} (arcsin (\frac{x}{3}))

\int \frac{4 x + 1}{4 x} d x

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