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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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