What is the integral of (3x^2-4x+3)/((x^2+1)(x-1)^2) ?

The integral of (3x^2-4x+3)/((x^2+1)(x-1)^2) is -1/(x-1)+2arctan(x)+C

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integral of (3x^2-4x+3)/((x^2+1)(x-1)^2)

\int \frac{3 x ^{2} - 4 x + 3}{( x ^{2} + 1 ) ( x - 1 ) ^{2}} d x

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

Solution steps

\int \frac{3 x ^{2} - 4 x + 3}{( x ^{2} + 1 ) ( x - 1 ) ^{2}} d x

Expand

\frac{3 x ^{2} - 4 x + 3}{( x ^{2} + 1 ) ( x - 1 ) ^{2}} : \frac{3 x ^{2}}{( x ^{2} + 1 ) ( x - 1 ) ^{2}} - \frac{4 x}{( x ^{2} + 1 ) ( x - 1 ) ^{2}} + \frac{3}{( x ^{2} + 1 ) ( x - 1 ) ^{2}}

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{3 x ^{2}}{( x ^{2} + 1 ) ( x - 1 ) ^{2}} d x - \int \frac{4 x}{( x ^{2} + 1 ) ( x - 1 ) ^{2}} d x + \int \frac{3}{( x ^{2} + 1 ) ( x - 1 ) ^{2}} d x

\int \frac{3 x ^{2}}{( x ^{2} + 1 ) ( x - 1 ) ^{2}} d x = 3 (- \frac{1}{4} ln x^{2} + 1 + \frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )})

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

\int \frac{3}{( x ^{2} + 1 ) ( x - 1 ) ^{2}} d x = 3 (\frac{1}{4} ln x^{2} + 1 - \frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )})

= 3 (- \frac{1}{4} ln x^{2} + 1 + \frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )}) - 4 (- \frac{1}{2} arctan (x) - \frac{1}{2 ( x - 1 )}) + 3 (\frac{1}{4} ln x^{2} + 1 - \frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )})

Simplify

3 (- \frac{1}{4} ln x^{2} + 1 + \frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )}) - 4 (- \frac{1}{2} arctan (x) - \frac{1}{2 ( x - 1 )}) + 3 (\frac{1}{4} ln x^{2} + 1 - \frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )}) : - \frac{1}{x - 1} + 2 arctan (x)

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

Add a constant to the solution

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

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What is the integral of (3x^2-4x+3)/((x^2+1)(x-1)^2) ?
The integral of (3x^2-4x+3)/((x^2+1)(x-1)^2) is -1/(x-1)+2arctan(x)+C