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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

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