What is the integral of (3x^3-x^2+6x+24)/(x^2-2x+5) ?

The integral of (3x^3-x^2+6x+24)/(x^2-2x+5) is (3x^2)/2+5x+3/2 ln|x^2-2x+5|-ln| 1/4 (x-1)^2+1|+C

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integral of (3x^3-x^2+6x+24)/(x^2-2x+5)

\int \frac{3 x ^{3} - x ^{2} + 6 x + 24}{x ^{2} - 2 x + 5} d x

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

Solution steps

\int \frac{3 x ^{3} - x ^{2} + 6 x + 24}{x ^{2} - 2 x + 5} d x

Expand

\frac{3 x ^{3} - x ^{2} + 6 x + 24}{x ^{2} - 2 x + 5} : \frac{3 x ^{3}}{x ^{2} - 2 x + 5} - \frac{x ^{2}}{x ^{2} - 2 x + 5} + \frac{6 x}{x ^{2} - 2 x + 5} + \frac{24}{x ^{2} - 2 x + 5}

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 ^{3}}{x ^{2} - 2 x + 5} d x - \int \frac{x ^{2}}{x ^{2} - 2 x + 5} d x + \int \frac{6 x}{x ^{2} - 2 x + 5} d x + \int \frac{24}{x ^{2} - 2 x + 5} d x

\int \frac{3 x ^{3}}{x ^{2} - 2 x + 5} d x = 3 (\frac{x ^{2}}{2} + 2 x - \frac{1}{2} ln x^{2} - 2 x + 5 - \frac{11}{2} arctan (\frac{1}{2} (x - 1)))

\int \frac{x ^{2}}{x ^{2} - 2 x + 5} d x = ln \frac{1}{4} (x - 1)^{2} + 1 - \frac{3}{2} arctan (\frac{1}{2} (x - 1)) + x

\int \frac{6 x}{x ^{2} - 2 x + 5} d x = 6 (\frac{1}{2} ln x^{2} - 2 x + 5 + \frac{1}{2} arctan (\frac{1}{2} (x - 1)))

\int \frac{24}{x ^{2} - 2 x + 5} d x = 12 arctan (\frac{1}{2} (x - 1))

= 3 (\frac{x ^{2}}{2} + 2 x - \frac{1}{2} ln x^{2} - 2 x + 5 - \frac{11}{2} arctan (\frac{1}{2} (x - 1))) - (ln \frac{1}{4} (x - 1)^{2} + 1 - \frac{3}{2} arctan (\frac{1}{2} (x - 1)) + x) + 6 (\frac{1}{2} ln x^{2} - 2 x + 5 + \frac{1}{2} arctan (\frac{1}{2} (x - 1))) + 12 arctan (\frac{1}{2} (x - 1))

Simplify

3 (\frac{x ^{2}}{2} + 2 x - \frac{1}{2} ln x^{2} - 2 x + 5 - \frac{11}{2} arctan (\frac{1}{2} (x - 1))) - (ln \frac{1}{4} (x - 1)^{2} + 1 - \frac{3}{2} arctan (\frac{1}{2} (x - 1)) + x) + 6 (\frac{1}{2} ln x^{2} - 2 x + 5 + \frac{1}{2} arctan (\frac{1}{2} (x - 1))) + 12 arctan (\frac{1}{2} (x - 1)) : \frac{3 x ^{2}}{2} + 5 x + \frac{3}{2} ln x^{2} - 2 x + 5 - ln \frac{1}{4} (x - 1)^{2} + 1

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

Add a constant to the solution

= \frac{3 x ^{2}}{2} + 5 x + \frac{3}{2} ln x^{2} - 2 x + 5 - ln \frac{1}{4} (x - 1)^{2} + 1 + C

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What is the integral of (3x^3-x^2+6x+24)/(x^2-2x+5) ?
The integral of (3x^3-x^2+6x+24)/(x^2-2x+5) is (3x^2)/2+5x+3/2 ln|x^2-2x+5|-ln| 1/4 (x-1)^2+1|+C