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

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

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integral of 1/((x-1)(x-3))

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

- \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{1}{2 ( x - 3 )} d x = \frac{1}{2} ln ∣ x - 3 ∣

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

Add a constant to the solution

= - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣ + C

\frac{d}{d x} (sin (\frac{3}{x}))

x \to 2 lim (\frac{x}{x + 2})

\frac{d}{d x} ((x^{2} + 1)^{3})

x \to - \infty lim (\frac{4 e ^{\frac{1}{x}} - 1}{1 + e ^{x}})

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