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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

ln x^{2} + 2 x + 5 - \frac{5}{2} arctan (\frac{1}{2} (x + 1)) + C

Solution steps

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

Expand

\frac{2 x - 3}{x ^{2} + 2 x + 5} : \frac{2 x}{x ^{2} + 2 x + 5} - \frac{3}{x ^{2} + 2 x + 5}

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

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

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

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

Add similar elements:

- arctan (\frac{1}{2} (x + 1)) - \frac{3}{2} arctan (\frac{1}{2} (x + 1)) = - \frac{5}{2} arctan (\frac{1}{2} (x + 1))

= ln x^{2} + 2 x + 5 - \frac{5}{2} arctan (\frac{1}{2} (x + 1))

Add a constant to the solution

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

\int - cos^{3} (t) d t

\int \frac{11 x + 8}{10 x ^{2} + 5 x} d x

\int \frac{1}{x ^{2} + R ^{2}} d x

\int 5 x^{4} e^{- 3 x} d x

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

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