# $\int \frac{1}{x(x+1)…(x+n)}dx$ [duplicate]

I have not managed to evaluate the integral. I feel the idea is just split the integral into a sum of integrals

$$\int \frac{1}{x(x+1)...(x+n)}dx = \int \left(\frac{A_1}{x} + \frac{A_2}{x+1} +... +\frac{A_n}{x+n} \right) dx$$

So, technically, the main problem is to determine the coefficients in the numerators of these fractions. Hhm, it has occurred to me that I might bypass the system of $$n$$ equations just putting $$x = 0$$, $$x = - 1$$,..., $$x = - n$$ after multipling the whole expression by $$x(x+1)...(x+n)$$. The idea can be of help to solve the risen problem.

Another idea is just to decompose the fraction by representing the numerator of the initial fraction, which is 1, as an expression considered to be suitable to attain the objective. For example,

$$1 = x + 1 - (x + 2) + (x + 3) - (x + 4) +...$$

I might attempt to adopt the idea but there is a nice trick that solves the problem? I came across the integral when searching for the competition problems on integration.

When trying to apply "brutal force" method, I got many brackets and equations afterwards. Hence, there must be more cunning technique.

• To expand in partial fraction, multiply the fraction by $x+k$, substitute $x=-k$, and you get two products in the LHS (on the left and on the right of the missing $x+k$ factor), and in the RHS there is just $A_k$. You may use factorials or binomials to simplify this. – Jean-Claude Arbaut Jan 16 at 14:05
• See solution in math.stackexchange.com/questions/3915792/… – Quanto Jan 16 at 16:21