What is the integral of (x^2-3)/((x^2+1)(x-1)) ?

The integral of (x^2-3)/((x^2+1)(x-1)) is ln|x^2+1|+2arctan(x)-ln|x-1|+C

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integral of (x^2-3)/((x^2+1)(x-1))

\int \frac{x ^{2} - 3}{( x ^{2} + 1 ) ( x - 1 )} d x

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

Solution steps

\int \frac{x ^{2} - 3}{( x ^{2} + 1 ) ( x - 1 )} d x

Expand

\frac{x ^{2} - 3}{( x ^{2} + 1 ) ( x - 1 )} : \frac{x ^{2}}{( x ^{2} + 1 ) ( x - 1 )} - \frac{3}{( 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{x ^{2}}{( x ^{2} + 1 ) ( x - 1 )} d x - \int \frac{3}{( x ^{2} + 1 ) ( x - 1 )} d x

\int \frac{x ^{2}}{( 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 ∣

\int \frac{3}{( x ^{2} + 1 ) ( x - 1 )} d x = 3 (\frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 - arctan (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 ∣ - 3 (\frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 - arctan (x)) + \frac{1}{2} ln ∣ x - 1 ∣)

Simplify

\frac{1}{2} (\frac{1}{2} ln x^{2} + 1 + arctan (x)) + \frac{1}{2} ln ∣ x - 1 ∣ - 3 (\frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 - arctan (x)) + \frac{1}{2} ln ∣ x - 1 ∣) : ln x^{2} + 1 + 2 arctan (x) - ln ∣ x - 1 ∣

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

Add a constant to the solution

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

What is the integral of (x^2-3)/((x^2+1)(x-1)) ?
The integral of (x^2-3)/((x^2+1)(x-1)) is ln|x^2+1|+2arctan(x)-ln|x-1|+C