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

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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

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