∫(∏r=0 to m (1/(x+r)))dx , find the value of this …

To evaluate the integral:

\[
\int \prod_{r=0}^{m} \frac{1}{x + r} \, dx
\]

we can proceed with the following steps:

Step 1: Express the Product as a Sum
The integrand is a product of terms of the form \(\frac{1}{x + r}\). To simplify the integration, we can use partial fraction decomposition. Assume that:

\[
\prod_{r=0}^{m} \frac{1}{x + r} = \sum_{r=0}^{m} \frac{A_r}{x + r}
\]

where \(A_r\) are constants to be determined.

Step 2: Determine the Constants \(A_r\)
Multiply both sides by \(\prod_{r=0}^{m} (x + r)\):

\[
1 = \sum_{r=0}^{m} A_r \prod_{\substack{k=0 \\ k \neq r}}^{m} (x + k)
\]

To find \(A_r\), set \(x = -r\). This eliminates all terms in the sum except the one corresponding to \(A_r\):

\[
1 = A_r \prod_{\substack{k=0 \\ k \neq r}}^{m} (-r + k)
\]

Simplify the product:

\[
A_r = \frac{1}{\prod_{\substack{k=0 \\ k \neq r}}^{m} (k – r)}
\]

This can be written as:

\[
A_r = \frac{(-1)^r}{r! (m – r)!}
\]

Step 3: Integrate Term by Term
Now, the integral becomes:

\[
\int \sum_{r=0}^{m} \frac{A_r}{x + r} \, dx = \sum_{r=0}^{m} A_r \int \frac{1}{x + r} \, dx
\]

The integral of \(\frac{1}{x + r}\) is \(\ln|x + r|\), so:

\[
\sum_{r=0}^{m} A_r \ln|x + r| + C
\]

Substitute \(A_r\):

\[
\sum_{r=0}^{m} \frac{(-1)^r}{r! (m – r)!} \ln|x + r| + C
\]

Step 4: Simplify the Expression
The sum can be written in terms of binomial coefficients:

\[
\sum_{r=0}^{m} \frac{(-1)^r}{r! (m – r)!} \ln|x + r| = \frac{1}{m!} \sum_{r=0}^{m} (-1)^r \binom{m}{r} \ln|x + r|
\]

Thus, the final result is:

\[
\boxed{\frac{1}{m!} \sum_{r=0}^{m} (-1)^r \binom{m}{r} \ln|x + r| + C}
\]