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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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