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

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

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integral of (x-2)/((x-3)(x-4))

\int \frac{x - 2}{( x - 3 ) ( x - 4 )} d x

2 ln ∣ x - 4 ∣ - ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{x - 2}{( x - 3 ) ( x - 4 )} d x

Expand

\frac{x - 2}{( x - 3 ) ( x - 4 )} : \frac{x}{( x - 3 ) ( x - 4 )} - \frac{2}{( x - 3 ) ( x - 4 )}

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{x}{( x - 3 ) ( x - 4 )} d x - \int \frac{2}{( x - 3 ) ( x - 4 )} d x

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

\int \frac{2}{( x - 3 ) ( x - 4 )} d x = 2 (- ln ∣ x - 3 ∣ + ln ∣ x - 4 ∣)

= 4 ln ∣ x - 4 ∣ - 3 ln ∣ x - 3 ∣ - 2 (- ln ∣ x - 3 ∣ + ln ∣ x - 4 ∣)

Simplify

4 ln ∣ x - 4 ∣ - 3 ln ∣ x - 3 ∣ - 2 (- ln ∣ x - 3 ∣ + ln ∣ x - 4 ∣) : 2 ln ∣ x - 4 ∣ - ln ∣ x - 3 ∣

= 2 ln ∣ x - 4 ∣ - ln ∣ x - 3 ∣

Add a constant to the solution

= 2 ln ∣ x - 4 ∣ - ln ∣ x - 3 ∣ + C

\int \frac{1}{t ^{3}} d t

x \to 1 + lim (\frac{1 + x}{x - 1})

x \to - \infty lim (5 x - 2^{x})

\frac{\partial}{\partial x} (\frac{9 y}{x ^{5}})

f (x) = 1 - 4 x

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