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

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

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integral of 1/((x-4)(x-1))

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

\frac{1}{3} ln ∣ x - 4 ∣ - \frac{1}{3} ln ∣ x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x - 4 ) ( x - 1 )} : \frac{1}{3 ( x - 4 )} - \frac{1}{3 ( x - 1 )}

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

\int \frac{1}{3 ( x - 4 )} d x = \frac{1}{3} ln ∣ x - 4 ∣

\int \frac{1}{3 ( x - 1 )} d x = \frac{1}{3} ln ∣ x - 1 ∣

= \frac{1}{3} ln ∣ x - 4 ∣ - \frac{1}{3} ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ x - 4 ∣ - \frac{1}{3} ln ∣ x - 1 ∣ + C

\int_{2}^{3} \frac{7}{3 - x} d x

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

x \to 0 lim (e^{a x})

\frac{t}{2 t - 9}

(6 x^{2} + 5)^{3}

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