What is the integral of ((2x+2))/((x^2+1)(x-1)^3) ?

The integral of ((2x+2))/((x^2+1)(x-1)^3) is arctan(x)+1/(x-1)-1/((x-1)^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of ((2x+2))/((x^2+1)(x-1)^3)

\int \frac{( 2 x + 2 )}{( x ^{2} + 1 ) ( x - 1 ) ^{3}} d x

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

Solution steps

\int \frac{2 x + 2}{( x ^{2} + 1 ) ( x - 1 ) ^{3}} d x

Take the partial fraction of

\frac{2 x + 2}{( x ^{2} + 1 ) ( x - 1 ) ^{3}} : \frac{1}{x ^{2} + 1} - \frac{1}{( x - 1 ) ^{2}} + \frac{2}{( x - 1 ) ^{3}}

= \int \frac{1}{x ^{2} + 1} - \frac{1}{( x - 1 ) ^{2}} + \frac{2}{( x - 1 ) ^{3}} 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 ^{2} + 1} d x - \int \frac{1}{( x - 1 ) ^{2}} d x + \int \frac{2}{( x - 1 ) ^{3}} d x

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

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

\int \frac{2}{( x - 1 ) ^{3}} d x = - \frac{1}{( x - 1 ) ^{2}}

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

Simplify

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

Add a constant to the solution

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

What is the integral of ((2x+2))/((x^2+1)(x-1)^3) ?
The integral of ((2x+2))/((x^2+1)(x-1)^3) is arctan(x)+1/(x-1)-1/((x-1)^2)+C