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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{4 u}{( u - 1 ) ^{3} - 6 ( u - 1 ) ^{2} + 8 u - 8} d u

= \int \frac{4 u}{( u - 1 ) ^{3} - 6 ( u - 1 ) ^{2} + 8 u - 8} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{u}{( u - 1 ) ^{3} - 6 ( u - 1 ) ^{2} + 8 u - 8} d u

Take the partial fraction of

\frac{u}{( u - 1 ) ^{3} - 6 ( u - 1 ) ^{2} + 8 u - 8} : \frac{1}{8 ( u - 1 )} - \frac{3}{4 ( u - 3 )} + \frac{5}{8 ( u - 5 )}

= 4 \cdot \int \frac{1}{8 ( u - 1 )} - \frac{3}{4 ( u - 3 )} + \frac{5}{8 ( u - 5 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int \frac{1}{8 ( u - 1 )} d u - \int \frac{3}{4 ( u - 3 )} d u + \int \frac{5}{8 ( u - 5 )} d u)

\int \frac{1}{8 ( u - 1 )} d u = \frac{1}{8} ln ∣ u - 1 ∣

\int \frac{3}{4 ( u - 3 )} d u = \frac{3}{4} ln ∣ u - 3 ∣

\int \frac{5}{8 ( u - 5 )} d u = \frac{5}{8} ln ∣ u - 5 ∣

= 4 (\frac{1}{8} ln ∣ u - 1 ∣ - \frac{3}{4} ln ∣ u - 3 ∣ + \frac{5}{8} ln ∣ u - 5 ∣)

Substitute back

u = 2 x + 1

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

Simplify

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

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

Add a constant to the solution

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

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