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

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

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

ln ∣ x ∣ + C

Solution steps

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

Simplify

\frac{x + 1}{x ^{2} + x} : \frac{1}{x}

= \int \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= ln ∣ x ∣

Add a constant to the solution

= ln ∣ x ∣ + C

\int \frac{1}{- x ^{2} + 4} d x

\int 3 x (3 x + 4)^{4} d x

\int \frac{1}{6 x ^{5}} d x

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

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

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