What is the integral of 4/(x^2-8x+12) ?

The integral of 4/(x^2-8x+12) is -ln|x-2|+ln|x-6|+C

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integral of 4/(x^2-8x+12)

\int \frac{4}{x ^{2} - 8 x + 12} d x

- ln ∣ x - 2 ∣ + ln ∣ x - 6 ∣ + C

Solution steps

\int \frac{4}{x ^{2} - 8 x + 12} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{x ^{2} - 8 x + 12} d x

Take the partial fraction of

\frac{1}{x ^{2} - 8 x + 12} : - \frac{1}{4 ( x - 2 )} + \frac{1}{4 ( x - 6 )}

= 4 \cdot \int - \frac{1}{4 ( x - 2 )} + \frac{1}{4 ( x - 6 )} 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

= 4 (- \int \frac{1}{4 ( x - 2 )} d x + \int \frac{1}{4 ( x - 6 )} d x)

\int \frac{1}{4 ( x - 2 )} d x = \frac{1}{4} ln ∣ x - 2 ∣

\int \frac{1}{4 ( x - 6 )} d x = \frac{1}{4} ln ∣ x - 6 ∣

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

Simplify

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

= - ln ∣ x - 2 ∣ + ln ∣ x - 6 ∣

Add a constant to the solution

= - ln ∣ x - 2 ∣ + ln ∣ x - 6 ∣ + C

y^{^{''}} + y^{^{'}} - 6 y = 4 e^{t}, y (0) = 0, y^{^{'}} (0) = 0

x + x^{- \frac{1}{3}}

\frac{\partial}{\partial x} (3 x^{2} y - 3 x + 3 y^{3})

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

(3 x^{2} - 3 x + 1) x^{3}

What is the integral of 4/(x^2-8x+12) ?
The integral of 4/(x^2-8x+12) is -ln|x-2|+ln|x-6|+C