What is the integral of (3x+1)/((x-4)(x-2)) ?

The integral of (3x+1)/((x-4)(x-2)) is 13/2 ln|x-4|-7/2 ln|x-2|+C

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integral of (3x+1)/((x-4)(x-2))

\int \frac{3 x + 1}{( x - 4 ) ( x - 2 )} d x

\frac{13}{2} ln ∣ x - 4 ∣ - \frac{7}{2} ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{3 x + 1}{( x - 4 ) ( x - 2 )} d x

Expand

\frac{3 x + 1}{( x - 4 ) ( x - 2 )} : \frac{3 x}{( x - 4 ) ( x - 2 )} + \frac{1}{( x - 4 ) ( x - 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 x}{( x - 4 ) ( x - 2 )} d x + \int \frac{1}{( x - 4 ) ( x - 2 )} d x

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

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

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

Simplify

3 (- ln ∣ x - 2 ∣ + 2 ln ∣ x - 4 ∣) + \frac{1}{2} ln ∣ x - 4 ∣ - \frac{1}{2} ln ∣ x - 2 ∣ : \frac{13}{2} ln ∣ x - 4 ∣ - \frac{7}{2} ln ∣ x - 2 ∣

= \frac{13}{2} ln ∣ x - 4 ∣ - \frac{7}{2} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{13}{2} ln ∣ x - 4 ∣ - \frac{7}{2} ln ∣ x - 2 ∣ + C

\int e^{c o s (27 t)} sin (27 t) d t

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

n = 0 \sum \infty \frac{x ^{n}}{n + 3}

\frac{d}{d y} (e^{x y^{2}})

\frac{d}{d x} (- \frac{2}{x})

What is the integral of (3x+1)/((x-4)(x-2)) ?
The integral of (3x+1)/((x-4)(x-2)) is 13/2 ln|x-4|-7/2 ln|x-2|+C