What is the integral of (x^2+1)/(x^3-x^2+3x+5) ?

The integral of (x^2+1)/(x^3-x^2+3x+5) is 1/4 ln|x+1|+1/8 (3ln|x^2-2x+5|+2arctan((x-1)/2))+C

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integral of (x^2+1)/(x^3-x^2+3x+5)

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

\frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{8} (3 ln x^{2} - 2 x + 5 + 2 arctan (\frac{x - 1}{2})) + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

\int \frac{3 x - 1}{4 ( x ^{2} - 2 x + 5 )} d x = \frac{1}{8} (3 ln x^{2} - 2 x + 5 + 2 arctan (\frac{x - 1}{2}))

= \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{8} (3 ln x^{2} - 2 x + 5 + 2 arctan (\frac{x - 1}{2}))

Add a constant to the solution

= \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{8} (3 ln x^{2} - 2 x + 5 + 2 arctan (\frac{x - 1}{2})) + C

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What is the integral of (x^2+1)/(x^3-x^2+3x+5) ?
The integral of (x^2+1)/(x^3-x^2+3x+5) is 1/4 ln|x+1|+1/8 (3ln|x^2-2x+5|+2arctan((x-1)/2))+C