What is the integral of (x^2+5x+10)/(x^3-x^2+4x-4) ?

The integral of (x^2+5x+10)/(x^3-x^2+4x-4) is 16/5 ln|x-1|+1/5 (-11/2 ln|x^2+4|+7arctan(x/2))+C

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integral of (x^2+5x+10)/(x^3-x^2+4x-4)

\int \frac{x ^{2} + 5 x + 10}{x ^{3} - x ^{2} + 4 x - 4} d x

\frac{16}{5} ln ∣ x - 1 ∣ + \frac{1}{5} (- \frac{11}{2} ln x^{2} + 4 + 7 arctan (\frac{x}{2})) + C

Solution steps

\int \frac{x ^{2} + 5 x + 10}{x ^{3} - x ^{2} + 4 x - 4} d x

Take the partial fraction of

\frac{x ^{2} + 5 x + 10}{x ^{3} - x ^{2} + 4 x - 4} : \frac{16}{5 ( x - 1 )} + \frac{- 11 x + 14}{5 ( x ^{2} + 4 )}

= \int \frac{16}{5 ( x - 1 )} + \frac{- 11 x + 14}{5 ( x ^{2} + 4 )} 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{16}{5 ( x - 1 )} d x + \int \frac{- 11 x + 14}{5 ( x ^{2} + 4 )} d x

\int \frac{16}{5 ( x - 1 )} d x = \frac{16}{5} ln ∣ x - 1 ∣

\int \frac{- 11 x + 14}{5 ( x ^{2} + 4 )} d x = \frac{1}{5} (- \frac{11}{2} ln x^{2} + 4 + 7 arctan (\frac{x}{2}))

= \frac{16}{5} ln ∣ x - 1 ∣ + \frac{1}{5} (- \frac{11}{2} ln x^{2} + 4 + 7 arctan (\frac{x}{2}))

Add a constant to the solution

= \frac{16}{5} ln ∣ x - 1 ∣ + \frac{1}{5} (- \frac{11}{2} ln x^{2} + 4 + 7 arctan (\frac{x}{2})) + C

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What is the integral of (x^2+5x+10)/(x^3-x^2+4x-4) ?
The integral of (x^2+5x+10)/(x^3-x^2+4x-4) is 16/5 ln|x-1|+1/5 (-11/2 ln|x^2+4|+7arctan(x/2))+C