What is the integral of 1/(x^2-5x+4) ?

The integral of 1/(x^2-5x+4) is -1/3 ln|x-1|+1/3 ln|x-4|+C

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integral of 1/(x^2-5x+4)

\int \frac{1}{x ^{2} - 5 x + 4} d x

- \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{1}{x ^{2} - 5 x + 4} d x

Take the partial fraction of

\frac{1}{x ^{2} - 5 x + 4} : - \frac{1}{3 ( x - 1 )} + \frac{1}{3 ( x - 4 )}

= \int - \frac{1}{3 ( x - 1 )} + \frac{1}{3 ( x - 4 )} 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}{3 ( x - 1 )} d x + \int \frac{1}{3 ( x - 4 )} d x

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

\int \frac{1}{3 ( x - 4 )} d x = \frac{1}{3} ln ∣ x - 4 ∣

= - \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x - 4 ∣

Add a constant to the solution

= - \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x - 4 ∣ + C

\int x x^{4} d x

y^{^{'}} + 5 y = t + e^{- 3 t}

\int \frac{x ^{3}}{8} d x

f (x) = e^{r x}

\frac{x ^{6}}{x ^{2}}

What is the integral of 1/(x^2-5x+4) ?
The integral of 1/(x^2-5x+4) is -1/3 ln|x-1|+1/3 ln|x-4|+C