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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

Add a constant to the solution

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

x \to - 6 lim (\frac{6 - ∣ x ∣}{6 + x})

\int (sec (x) + tan (x))^{2} d x

\frac{\partial}{\partial y} (2 x + y cos (x y))

\int_{0}^{2} 17 d x

e^{\frac{z}{z - 3}}

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