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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

\int_{0}^{π} 7 e^{x} sin (x) d x

f (x) = \frac{1}{3 + 2 x}, at x = 2

\int \frac{x ^{2}}{1 + x ^{4}} d x

\frac{d}{d x} (\frac{cos ( sin ( x + 1 ) )}{tan ( \frac{1}{x} )})

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