What is the integral of 1/((x^2-6x+5)) ?

The integral of 1/((x^2-6x+5)) is -1/4 ln|x-1|+1/4 ln|x-5|+C

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integral of 1/((x^2-6x+5))

integral

\frac{1}{( x ^{2} - 6 x + 5 )}

- \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ x - 5 ∣ + C

Solution steps

\int \frac{1}{x ^{2} - 6 x + 5} d x

Take the partial fraction of

\frac{1}{x ^{2} - 6 x + 5} : - \frac{1}{4 ( x - 1 )} + \frac{1}{4 ( x - 5 )}

= \int - \frac{1}{4 ( x - 1 )} + \frac{1}{4 ( 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}{4 ( x - 1 )} d x + \int \frac{1}{4 ( x - 5 )} d x

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

\int \frac{1}{4 ( x - 5 )} d x = \frac{1}{4} ln ∣ x - 5 ∣

= - \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ x - 5 ∣

Add a constant to the solution

= - \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ x - 5 ∣ + C

\frac{5}{x + 1}

\frac{( 1 + x ) ^{2}}{x ^{2}}

\frac{( 1 - x ^{2} )}{( e ^{2} x )}

\frac{x ^{- 2}}{3}

sin (2 x) + cos (x)

What is the integral of 1/((x^2-6x+5)) ?
The integral of 1/((x^2-6x+5)) is -1/4 ln|x-1|+1/4 ln|x-5|+C