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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(x+1)x

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

Simplify

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

Add a constant to the solution

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

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