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

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

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

- ln x^{2} + 1 + 2 ln ∣ x - 1 ∣ + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

= - ln x^{2} + 1 + 2 ln ∣ x - 1 ∣

Add a constant to the solution

= - ln x^{2} + 1 + 2 ln ∣ x - 1 ∣ + C

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

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

\int \frac{1}{4 x ^{3} + 4 x ^{2} + x} d x

\int 10 e^{- x} d x

\int \frac{1}{3 cos ( x )} d x

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