What is the integral of (3x^2-2x+19)/((x-1)(x^2+9)) ?

The integral of (3x^2-2x+19)/((x-1)(x^2+9)) is 2ln|x-1|+1/2 ln|x^2+9|-1/3 arctan(x/3)+C

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integral of (3x^2-2x+19)/((x-1)(x^2+9))

\int \frac{3 x ^{2} - 2 x + 19}{( x - 1 ) ( x ^{2} + 9 )} d x

2 ln ∣ x - 1 ∣ + \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}) + C

Solution steps

\int \frac{3 x ^{2} - 2 x + 19}{( x - 1 ) ( x ^{2} + 9 )} d x

Take the partial fraction of

\frac{3 x ^{2} - 2 x + 19}{( x - 1 ) ( x ^{2} + 9 )} : \frac{2}{x - 1} + \frac{x - 1}{x ^{2} + 9}

= \int \frac{2}{x - 1} + \frac{x - 1}{x ^{2} + 9} 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{2}{x - 1} d x + \int \frac{x - 1}{x ^{2} + 9} d x

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

\int \frac{x - 1}{x ^{2} + 9} d x = \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3})

= 2 ln ∣ x - 1 ∣ + \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3})

Add a constant to the solution

= 2 ln ∣ x - 1 ∣ + \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}) + C

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What is the integral of (3x^2-2x+19)/((x-1)(x^2+9)) ?
The integral of (3x^2-2x+19)/((x-1)(x^2+9)) is 2ln|x-1|+1/2 ln|x^2+9|-1/3 arctan(x/3)+C