What is the integral of (2x^2-3x-3)/(x^3-3x^2+7x-5) ?

The integral of (2x^2-3x-3)/(x^3-3x^2+7x-5) is -ln|x-1|+3/2 ln|x^2-2x+5|+1/2 arctan(1/2 (x-1))+C

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integral of (2x^2-3x-3)/(x^3-3x^2+7x-5)

\int \frac{2 x ^{2} - 3 x - 3}{x ^{3} - 3 x ^{2} + 7 x - 5} d x

- ln ∣ x - 1 ∣ + \frac{3}{2} ln x^{2} - 2 x + 5 + \frac{1}{2} arctan (\frac{1}{2} (x - 1)) + C

Solution steps

\int \frac{2 x ^{2} - 3 x - 3}{x ^{3} - 3 x ^{2} + 7 x - 5} d x

Factor

\frac{2 x ^{2} - 3 x - 3}{x ^{3} - 3 x ^{2} + 7 x - 5}

= \int \frac{2 x ^{2} - 3 x - 3}{( x - 1 ) ( x ^{2} - 2 x + 5 )} d x

Apply u-substitution

= \int \frac{2 u ^{2} + u - 4}{u ( u ^{2} + 4 )} d u

Take the partial fraction of

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

= \int - \frac{1}{u} + \frac{3 u + 1}{u ^{2} + 4} d u

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{1}{u} d u + \int \frac{3 u + 1}{u ^{2} + 4} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{3 u + 1}{u ^{2} + 4} d u = \frac{3}{2} ln u^{2} + 4 + \frac{1}{2} arctan (\frac{u}{2})

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

Substitute back

u = x - 1

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

Simplify

- ln ∣ x - 1 ∣ + \frac{3}{2} ln (x - 1)^{2} + 4 + \frac{1}{2} arctan (\frac{x - 1}{2}) : - ln ∣ x - 1 ∣ + \frac{3}{2} ln x^{2} - 2 x + 5 + \frac{1}{2} arctan (\frac{1}{2} (x - 1))

= - ln ∣ x - 1 ∣ + \frac{3}{2} ln x^{2} - 2 x + 5 + \frac{1}{2} arctan (\frac{1}{2} (x - 1))

Add a constant to the solution

= - ln ∣ x - 1 ∣ + \frac{3}{2} ln x^{2} - 2 x + 5 + \frac{1}{2} arctan (\frac{1}{2} (x - 1)) + C

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What is the integral of (2x^2-3x-3)/(x^3-3x^2+7x-5) ?
The integral of (2x^2-3x-3)/(x^3-3x^2+7x-5) is -ln|x-1|+3/2 ln|x^2-2x+5|+1/2 arctan(1/2 (x-1))+C