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

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

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integral of 1/((x-6)(x-2))

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

\frac{1}{4} ln ∣ x - 6 ∣ - \frac{1}{4} ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

f (x) = 4 sec (x), at \frac{π}{3}

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

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = - 4, y^{^{'}} (0) = - 1

e^{x} - 1

\int_{1}^{\infty} x^{- \frac{5}{4}} d x

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