What is the integral of 1/((x-5)(x-2)) ?

The integral of 1/((x-5)(x-2)) is 1/3 ln|x-5|-1/3 ln|x-2|+C

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integral of 1/((x-5)(x-2))

\int \frac{1}{( x - 5 ) ( x - 2 )} d x

\frac{1}{3} ln ∣ x - 5 ∣ - \frac{1}{3} ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{1}{( x - 5 ) ( x - 2 )} d x

Take the partial fraction of

\frac{1}{( x - 5 ) ( x - 2 )} : \frac{1}{3 ( x - 5 )} - \frac{1}{3 ( x - 2 )}

= \int \frac{1}{3 ( x - 5 )} - \frac{1}{3 ( 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}{3 ( x - 5 )} d x - \int \frac{1}{3 ( x - 2 )} d x

\int \frac{1}{3 ( x - 5 )} d x = \frac{1}{3} ln ∣ x - 5 ∣

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

= \frac{1}{3} ln ∣ x - 5 ∣ - \frac{1}{3} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ x - 5 ∣ - \frac{1}{3} ln ∣ x - 2 ∣ + C

\frac{\partial}{\partial x} ((e^{x} + e^{y})^{5})

\int_{1}^{2} (x - 1) d x

y^{^{'}} = 3 cos (t) - 2 y

\frac{3}{x + 2}

\int_{1}^{3} (4 x^{3} - 3 x^{2}) d x

What is the integral of 1/((x-5)(x-2)) ?
The integral of 1/((x-5)(x-2)) is 1/3 ln|x-5|-1/3 ln|x-2|+C