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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

ln x^{2} - 3 x + 5 + C

Solution steps

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

Apply u-substitution

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = x^{2} - 3 x + 5

= ln x^{2} - 3 x + 5

Add a constant to the solution

= ln x^{2} - 3 x + 5 + C

\int \frac{1}{t ^{2} + 81} d t

\int (1 + cos (2 x))^{3} d x

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

\int sin (\frac{x}{6}) d x

\int (x + 11) 22 x + x^{2} d x

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