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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

ln 1 + y^{2} (- 1 + x) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

\int \frac{y}{1 + y ^{2}} d y = \frac{1}{2} ln 1 + y^{2}

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

Simplify

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

= ln 1 + y^{2} (- 1 + x)

Add a constant to the solution

= ln 1 + y^{2} (- 1 + x) + C

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