What is the integral of (2x+3)/(x^2-5x+4) ?

The integral of (2x+3)/(x^2-5x+4) is -5/3 ln|x-1|+11/3 ln|x-4|+C

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integral of (2x+3)/(x^2-5x+4)

\int \frac{2 x + 3}{x ^{2} - 5 x + 4} d x

- \frac{5}{3} ln ∣ x - 1 ∣ + \frac{11}{3} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{2 x + 3}{x ^{2} - 5 x + 4} d x

Take the partial fraction of

\frac{2 x + 3}{x ^{2} - 5 x + 4} : - \frac{5}{3 ( x - 1 )} + \frac{11}{3 ( x - 4 )}

= \int - \frac{5}{3 ( x - 1 )} + \frac{11}{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{5}{3 ( x - 1 )} d x + \int \frac{11}{3 ( x - 4 )} d x

\int \frac{5}{3 ( x - 1 )} d x = \frac{5}{3} ln ∣ x - 1 ∣

\int \frac{11}{3 ( x - 4 )} d x = \frac{11}{3} ln ∣ x - 4 ∣

= - \frac{5}{3} ln ∣ x - 1 ∣ + \frac{11}{3} ln ∣ x - 4 ∣

Add a constant to the solution

= - \frac{5}{3} ln ∣ x - 1 ∣ + \frac{11}{3} ln ∣ x - 4 ∣ + C

\int \frac{1}{6} x d x

\frac{x}{3}

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

f (x) = x - 4

8 x + \frac{d y}{d x} = 0

What is the integral of (2x+3)/(x^2-5x+4) ?
The integral of (2x+3)/(x^2-5x+4) is -5/3 ln|x-1|+11/3 ln|x-4|+C