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

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

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

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

Solution steps

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

Take the partial fraction of

\frac{3 x + 1}{x ( x + 1 ) ^{2}} : \frac{1}{x} - \frac{1}{x + 1} + \frac{2}{( x + 1 ) ^{2}}

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

e^{- 2 t - 5}

\frac{\partial}{\partial x} (y \cdot ln (x))

\frac{d}{d x} (s e^{- s x y (s, x)})

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

\int_{0}^{π} sin^{2} (x) d x

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