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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

\int \frac{1 + 3 ln ( x )}{x} d x

\int \frac{( x ^{3} - 8 )}{( x - 2 )} d x

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

\int \frac{x ^{3}}{1 - 4 x ^{8}} d x

\int (5 x^{3} + 1)^{4} x^{5} d x

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