What is the integral of (-4x+8)/((x^2+1)(x-1)) ?

The integral of (-4x+8)/((x^2+1)(x-1)) is -ln|x^2+1|-6arctan(x)+2ln|x-1|+C

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integral of (-4x+8)/((x^2+1)(x-1))

\int \frac{- 4 x + 8}{( x ^{2} + 1 ) ( x - 1 )} d x

- ln x^{2} + 1 - 6 arctan (x) + 2 ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{- 4 x + 8}{( x ^{2} + 1 ) ( x - 1 )} d x

Expand

\frac{- 4 x + 8}{( x ^{2} + 1 ) ( x - 1 )} : - \frac{4 x}{( x ^{2} + 1 ) ( x - 1 )} + \frac{8}{( x ^{2} + 1 ) ( x - 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{4 x}{( x ^{2} + 1 ) ( x - 1 )} d x + \int \frac{8}{( x ^{2} + 1 ) ( x - 1 )} d x

\int \frac{4 x}{( x ^{2} + 1 ) ( x - 1 )} d x = 4 (\frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 + arctan (x)) + \frac{1}{2} ln ∣ x - 1 ∣)

\int \frac{8}{( x ^{2} + 1 ) ( x - 1 )} d x = 8 (\frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 - arctan (x)) + \frac{1}{2} ln ∣ x - 1 ∣)

= - 4 (\frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 + arctan (x)) + \frac{1}{2} ln ∣ x - 1 ∣) + 8 (\frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 - arctan (x)) + \frac{1}{2} ln ∣ x - 1 ∣)

Simplify

- 4 (\frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 + arctan (x)) + \frac{1}{2} ln ∣ x - 1 ∣) + 8 (\frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 - arctan (x)) + \frac{1}{2} ln ∣ x - 1 ∣) : - ln x^{2} + 1 - 6 arctan (x) + 2 ln ∣ x - 1 ∣

= - ln x^{2} + 1 - 6 arctan (x) + 2 ln ∣ x - 1 ∣

Add a constant to the solution

= - ln x^{2} + 1 - 6 arctan (x) + 2 ln ∣ x - 1 ∣ + C

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What is the integral of (-4x+8)/((x^2+1)(x-1)) ?
The integral of (-4x+8)/((x^2+1)(x-1)) is -ln|x^2+1|-6arctan(x)+2ln|x-1|+C