What is the integral of (5x+8)/(x^2+4x+4) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (5x+8)/(x^2+4x+4)

\int \frac{5 x + 8}{x ^{2} + 4 x + 4} d x

5 ln ∣ x + 2 ∣ + \frac{2}{x + 2} + C

Solution steps

\int \frac{5 x + 8}{x ^{2} + 4 x + 4} d x

Expand

\frac{5 x + 8}{x ^{2} + 4 x + 4} : \frac{5 x}{x ^{2} + 4 x + 4} + \frac{8}{x ^{2} + 4 x + 4}

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{5 x}{x ^{2} + 4 x + 4} d x + \int \frac{8}{x ^{2} + 4 x + 4} d x

\int \frac{5 x}{x ^{2} + 4 x + 4} d x = 5 (ln ∣ x + 2 ∣ + \frac{2}{x + 2})

\int \frac{8}{x ^{2} + 4 x + 4} d x = - \frac{8}{x + 2}

= 5 (ln ∣ x + 2 ∣ + \frac{2}{x + 2}) - \frac{8}{x + 2}

Simplify

5 (ln ∣ x + 2 ∣ + \frac{2}{x + 2}) - \frac{8}{x + 2} : 5 ln ∣ x + 2 ∣ + \frac{2}{x + 2}

= 5 ln ∣ x + 2 ∣ + \frac{2}{x + 2}

Add a constant to the solution

= 5 ln ∣ x + 2 ∣ + \frac{2}{x + 2} + C

x \to \infty lim ((1 + \frac{1}{2 x})^{3 x})

\int_{2}^{6} t^{3} ln (4 t) d t

y^{^{'}} = 5 y sin (4 x), y (\frac{π}{8}) = 3

n = 1 \sum \infty \frac{1}{π ^{n}}

x \to 0 lim (x (\frac{1}{2 x}))

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