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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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