What is the integral of (x+1)/(x^2-6x+8) ?

The integral of (x+1)/(x^2-6x+8) is 1/2 ln|x^2-6x+8|-4(1/2 ln|x-2|-1/2 ln|x-4|)+C

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integral of (x+1)/(x^2-6x+8)

\int \frac{x + 1}{x ^{2} - 6 x + 8} d x

\frac{1}{2} ln x^{2} - 6 x + 8 - 4 (\frac{1}{2} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 4 ∣) + C

Solution steps

\int \frac{x + 1}{x ^{2} - 6 x + 8} d x

Complete the square

x^{2} - 6 x + 8 : (x - 3)^{2} - 1

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

Apply u-substitution

: \int \frac{u + 4}{u ^{2} - 1} d u

= \int \frac{u + 4}{u ^{2} - 1} d u

Expand

\frac{u + 4}{u ^{2} - 1} : \frac{u}{u ^{2} - 1} + \frac{4}{u ^{2} - 1}

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{u}{u ^{2} - 1} d u + \int \frac{4}{u ^{2} - 1} d u

\int \frac{u}{u ^{2} - 1} d u = \frac{1}{2} ln u^{2} - 1

\int \frac{4}{u ^{2} - 1} d u = - 4 (\frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2} ln ∣ u - 1 ∣)

= \frac{1}{2} ln u^{2} - 1 - 4 (\frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2} ln ∣ u - 1 ∣)

Substitute back

u = x - 3

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

Simplify

\frac{1}{2} ln (x - 3)^{2} - 1 - 4 (\frac{1}{2} ln ∣ x - 3 + 1 ∣ - \frac{1}{2} ln ∣ x - 3 - 1 ∣) : \frac{1}{2} ln x^{2} - 6 x + 8 - 4 (\frac{1}{2} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 4 ∣)

= \frac{1}{2} ln x^{2} - 6 x + 8 - 4 (\frac{1}{2} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 4 ∣)

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 6 x + 8 - 4 (\frac{1}{2} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 4 ∣) + C

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What is the integral of (x+1)/(x^2-6x+8) ?
The integral of (x+1)/(x^2-6x+8) is 1/2 ln|x^2-6x+8|-4(1/2 ln|x-2|-1/2 ln|x-4|)+C