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)

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

- \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

d er i v a t i v e 7 (x^{2} + 3 x)^{6} (2 x + 3)

\int e^{\frac{x}{60}} d x

d er i v a t i v e x^{4} a^{5} + y^{3} a + z^{6}

y^{^{''}} + y = x cos^{3} (x)

d er i v a t i v e sin^{2} (7 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