What is the integral of arctan(1-x) ?

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

\int arctan (1 - x) d x

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

Solution steps

\int arctan (1 - x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

\int \frac{e ^{- y}}{9 - 16 e ^{- 2 y}} d y

\int (2 x - 1) cos (5 x) d x

\int (\frac{( ln ( x ) ) ^{5}}{x}) d x

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

\int 2 yz - 3 d y

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