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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

- \frac{2}{x + 1} + ln ∣ x + 1 ∣ + C

Solution steps

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

Expand

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

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

\int \frac{x}{x ^{2} + 2 x + 1} d x = ln ∣ x + 1 ∣ + \frac{1}{x + 1}

= - \frac{3}{x + 1} + ln ∣ x + 1 ∣ + \frac{1}{x + 1}

Simplify

- \frac{3}{x + 1} + ln ∣ x + 1 ∣ + \frac{1}{x + 1} : - \frac{2}{x + 1} + ln ∣ x + 1 ∣

= - \frac{2}{x + 1} + ln ∣ x + 1 ∣

Add a constant to the solution

= - \frac{2}{x + 1} + ln ∣ x + 1 ∣ + C

\int \frac{sin ( x )}{( cos ( x ) ) ^{4}} d x

\int e^{- 0.6 t} d t

\int (\frac{x}{16 - x ^{2}}) d x

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

\int e^{\frac{x ^{2} - 2 x}{100}} (20 x - 20) d x

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