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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

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

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

Take the partial fraction of

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

= 2 \cdot \int \frac{1}{u - 1} - \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

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

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

\int \frac{1}{u} d u = ln ∣ u ∣

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

= 2 (ln ∣ u - 1 ∣ - ln ∣ u ∣ - (- \frac{1}{u}))

Substitute back

u = x + 1

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

Simplify

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

Add a constant to the solution

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

\int \frac{1}{x 1 + 4 x} d x

\int 5 x^{4} sinh (x^{5}) d x

\int arcsin (2 y) d y

\int \frac{10 x}{3 + 5 x ^{2}} d x

\int \frac{x}{x ^{4} - 2 x ^{2} + 10} d x

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