What is the integral of (x-1)/(x^2-3x+1) ?

The integral of (x-1)/(x^2-3x+1) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x-1)/(x^2-3x+1)

\int \frac{x - 1}{x ^{2} - 3 x + 1} d x

\frac{1}{2} ln 4 x^{2} - 12 x + 4 - \frac{1}{2 5} ln \frac{1}{5} (2 x - 3) + 1 + \frac{1}{2 5} ln \frac{1}{5} (2 x - 3) - 1 + C

Solution steps

\int \frac{x - 1}{x ^{2} - 3 x + 1} d x

Expand

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

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{x}{x ^{2} - 3 x + 1} d x - \int \frac{1}{x ^{2} - 3 x + 1} d x

\int \frac{x}{x ^{2} - 3 x + 1} d x = 2 (\frac{1}{4} ln 4 x^{2} - 12 x + 4 - \frac{3}{4 5} (ln \frac{2 x - 3}{5} + 1 - ln \frac{2 x - 3}{5} - 1))

\int \frac{1}{x ^{2} - 3 x + 1} d x = - \frac{1}{5} (ln \frac{2 x - 3}{5} + 1 - ln \frac{2 x - 3}{5} - 1)

= 2 (\frac{1}{4} ln 4 x^{2} - 12 x + 4 - \frac{3}{4 5} (ln \frac{2 x - 3}{5} + 1 - ln \frac{2 x - 3}{5} - 1)) - (- \frac{1}{5} (ln \frac{2 x - 3}{5} + 1 - ln \frac{2 x - 3}{5} - 1))

Simplify

2 (\frac{1}{4} ln 4 x^{2} - 12 x + 4 - \frac{3}{4 5} (ln \frac{2 x - 3}{5} + 1 - ln \frac{2 x - 3}{5} - 1)) - (- \frac{1}{5} (ln \frac{2 x - 3}{5} + 1 - ln \frac{2 x - 3}{5} - 1)) : \frac{1}{2} ln 4 x^{2} - 12 x + 4 - \frac{1}{2 5} ln \frac{1}{5} (2 x - 3) + 1 + \frac{1}{2 5} ln \frac{1}{5} (2 x - 3) - 1

= \frac{1}{2} ln 4 x^{2} - 12 x + 4 - \frac{1}{2 5} ln \frac{1}{5} (2 x - 3) + 1 + \frac{1}{2 5} ln \frac{1}{5} (2 x - 3) - 1

Add a constant to the solution

= \frac{1}{2} ln 4 x^{2} - 12 x + 4 - \frac{1}{2 5} ln \frac{1}{5} (2 x - 3) + 1 + \frac{1}{2 5} ln \frac{1}{5} (2 x - 3) - 1 + C

\int 3 x (x^{2} - 5)^{5} d x

\int (- 3 + \frac{12}{p}) d p

\int 4 sin (x^{2}) d x

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

\int \frac{2}{x ln ( 9 x )} d x