What is the integral of (x-1)/(4x^2+4x+5) ?

The integral of (x-1)/(4x^2+4x+5) is 1/8 ln|4x^2+4x+5|-3/8 arctan(x+1/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x-1)/(4x^2+4x+5)

\int \frac{x - 1}{4 x ^{2} + 4 x + 5} d x

\frac{1}{8} ln 4 x^{2} + 4 x + 5 - \frac{3}{8} arctan (x + \frac{1}{2}) + C

Solution steps

\int \frac{x - 1}{4 x ^{2} + 4 x + 5} d x

Expand

\frac{x - 1}{4 x ^{2} + 4 x + 5} : \frac{x}{4 x ^{2} + 4 x + 5} - \frac{1}{4 x ^{2} + 4 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{x}{4 x ^{2} + 4 x + 5} d x - \int \frac{1}{4 x ^{2} + 4 x + 5} d x

\int \frac{x}{4 x ^{2} + 4 x + 5} d x = \frac{1}{8} (ln 4 x^{2} + 4 x + 5 - arctan (x + \frac{1}{2}))

\int \frac{1}{4 x ^{2} + 4 x + 5} d x = \frac{1}{4} arctan (x + \frac{1}{2})

= \frac{1}{8} (ln 4 x^{2} + 4 x + 5 - arctan (x + \frac{1}{2})) - \frac{1}{4} arctan (x + \frac{1}{2})

Simplify

\frac{1}{8} (ln 4 x^{2} + 4 x + 5 - arctan (x + \frac{1}{2})) - \frac{1}{4} arctan (x + \frac{1}{2}) : \frac{1}{8} ln 4 x^{2} + 4 x + 5 - \frac{3}{8} arctan (x + \frac{1}{2})

= \frac{1}{8} ln 4 x^{2} + 4 x + 5 - \frac{3}{8} arctan (x + \frac{1}{2})

Add a constant to the solution

= \frac{1}{8} ln 4 x^{2} + 4 x + 5 - \frac{3}{8} arctan (x + \frac{1}{2}) + C

What is the integral of (x-1)/(4x^2+4x+5) ?
The integral of (x-1)/(4x^2+4x+5) is 1/8 ln|4x^2+4x+5|-3/8 arctan(x+1/2)+C