What is the integral of ((-2x+4))/((x^2+1)(x-1)^2) ?

The integral of ((-2x+4))/((x^2+1)(x-1)^2) is ln|x^2+1|+arctan(x)-2ln|x-1|-1/(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+4))/((x^2+1)(x-1)^2)

\int \frac{( - 2 x + 4 )}{( x ^{2} + 1 ) ( x - 1 ) ^{2}} d x

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

Solution steps

\int \frac{- 2 x + 4}{( x ^{2} + 1 ) ( x - 1 ) ^{2}} d x

Take the partial fraction of

\frac{- 2 x + 4}{( x ^{2} + 1 ) ( x - 1 ) ^{2}} : \frac{2 x + 1}{x ^{2} + 1} - \frac{2}{x - 1} + \frac{1}{( x - 1 ) ^{2}}

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

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

\int \frac{2}{x - 1} d x = 2 ln ∣ x - 1 ∣

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

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

Add a constant to the solution

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

\frac{d}{d x} (x^{2 s i n (x)})

f (x) = 2 π x^{2}

f (x) = 2 x^{2} ln (2 x)

\frac{\partial}{\partial x} ((ln (x^{2} + y^{2}))^{\frac{1}{2}})

\int (x - 2 y) d y

What is the integral of ((-2x+4))/((x^2+1)(x-1)^2) ?
The integral of ((-2x+4))/((x^2+1)(x-1)^2) is ln|x^2+1|+arctan(x)-2ln|x-1|-1/(x-1)+C