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

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

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

ln ∣ x + 1 ∣ + \frac{1}{x + 1} + C

Solution steps

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

Complete the square

x^{2} + 2 x + 1 : (x + 1)^{2}

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

Apply u-substitution

: \int \frac{u - 1}{u ^{2}} d u

= \int \frac{u - 1}{u ^{2}} d u

Expand

\frac{u - 1}{u ^{2}} : \frac{1}{u} - \frac{1}{u ^{2}}

= \int \frac{1}{u} - \frac{1}{u ^{2}} d u

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{1}{u} d u - \int \frac{1}{u ^{2}} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

= ln ∣ u ∣ - (- \frac{1}{u})

Substitute back

u = x + 1

= ln ∣ x + 1 ∣ - (- \frac{1}{x + 1})

Simplify

= ln ∣ x + 1 ∣ + \frac{1}{x + 1}

Add a constant to the solution

= ln ∣ x + 1 ∣ + \frac{1}{x + 1} + C

\frac{d}{d x} (2 x^{2} - 4)

\frac{d}{d x} (ln x^{2} - 4)

f (x) = - 2 x^{2} - 12 x - 4

\frac{t d y}{d t} = t^{3} + 8 t^{3} y

\frac{d}{d x} (\frac{( x + 3 ) ^{2}}{e ^{x} \cdot cos ( x )})

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