What is the integral of (x^2-4x+7)/(x^3-x^2+x+3) ?

The integral of (x^2-4x+7)/(x^3-x^2+x+3) is 2ln|x+1|-1/2 ln|x^2-2x+3|+C

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integral of (x^2-4x+7)/(x^3-x^2+x+3)

\int \frac{x ^{2} - 4 x + 7}{x ^{3} - x ^{2} + x + 3} d x

2 ln ∣ x + 1 ∣ - \frac{1}{2} ln x^{2} - 2 x + 3 + C

Solution steps

\int \frac{x ^{2} - 4 x + 7}{x ^{3} - x ^{2} + x + 3} d x

Take the partial fraction of

\frac{x ^{2} - 4 x + 7}{x ^{3} - x ^{2} + x + 3} : \frac{2}{x + 1} + \frac{- x + 1}{x ^{2} - 2 x + 3}

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

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

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

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

Add similar elements:

- \frac{1}{2} arctan (\frac{1}{2} (x - 1)) + \frac{1}{2} arctan (\frac{1}{2} (x - 1)) = 0

= 2 ln ∣ x + 1 ∣ - \frac{1}{2} ln x^{2} - 2 x + 3

Add a constant to the solution

= 2 ln ∣ x + 1 ∣ - \frac{1}{2} ln x^{2} - 2 x + 3 + C

x \to 0 + lim (ln (x + 1))

y y^{^{'}} = x

(0, 6), (3, - 24)

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

\int 1 + 2 x d x

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