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

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

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integral of (3w-7w^2+7)/(w^3-3w+2)

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

- 4 ln ∣ w - 1 ∣ - 3 ln ∣ w + 2 ∣ - \frac{1}{w - 1} + C

Solution steps

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

Expand

\frac{3 w - 7 w ^{2} + 7}{w ^{3} - 3 w + 2} : \frac{3 w}{w ^{3} - 3 w + 2} - \frac{7 w ^{2}}{w ^{3} - 3 w + 2} + \frac{7}{w ^{3} - 3 w + 2}

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

\int \frac{3 w}{w ^{3} - 3 w + 2} d w = \frac{2}{3} ln ∣ w - 1 ∣ - \frac{1}{w - 1} - \frac{2}{3} ln ∣ w + 2 ∣

\int \frac{7 w ^{2}}{w ^{3} - 3 w + 2} d w = 7 (\frac{5}{9} ln ∣ w - 1 ∣ - \frac{1}{3 ( w - 1 )} + \frac{4}{9} ln ∣ w + 2 ∣)

\int \frac{7}{w ^{3} - 3 w + 2} d w = 7 (- \frac{1}{9} ln ∣ w - 1 ∣ - \frac{1}{3 ( w - 1 )} + \frac{1}{9} ln ∣ w + 2 ∣)

= \frac{2}{3} ln ∣ w - 1 ∣ - \frac{1}{w - 1} - \frac{2}{3} ln ∣ w + 2 ∣ - 7 (\frac{5}{9} ln ∣ w - 1 ∣ - \frac{1}{3 ( w - 1 )} + \frac{4}{9} ln ∣ w + 2 ∣) + 7 (- \frac{1}{9} ln ∣ w - 1 ∣ - \frac{1}{3 ( w - 1 )} + \frac{1}{9} ln ∣ w + 2 ∣)

Simplify

\frac{2}{3} ln ∣ w - 1 ∣ - \frac{1}{w - 1} - \frac{2}{3} ln ∣ w + 2 ∣ - 7 (\frac{5}{9} ln ∣ w - 1 ∣ - \frac{1}{3 ( w - 1 )} + \frac{4}{9} ln ∣ w + 2 ∣) + 7 (- \frac{1}{9} ln ∣ w - 1 ∣ - \frac{1}{3 ( w - 1 )} + \frac{1}{9} ln ∣ w + 2 ∣) : - 4 ln ∣ w - 1 ∣ - 3 ln ∣ w + 2 ∣ - \frac{1}{w - 1}

= - 4 ln ∣ w - 1 ∣ - 3 ln ∣ w + 2 ∣ - \frac{1}{w - 1}

Add a constant to the solution

= - 4 ln ∣ w - 1 ∣ - 3 ln ∣ w + 2 ∣ - \frac{1}{w - 1} + C

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