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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{4}{x - 1} d x = 4 ln ∣ x - 1 ∣

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

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

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

Simplify

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

Add a constant to the solution

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

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