This problem requires evaluating an indefinite integral. We need to find the value of the integral:

\(\rm \int \frac{(x+1)}{x(xe^x + 1)}\ dx\)

We will use substitution and partial fraction decomposition to solve this integral.

Step 1: Applying Substitution

Let's choose a substitution that simplifies the integrand. Consider the term \(xe^x + 1\). Let's substitute:

\(u = xe^x + 1\)

Now, we need to find the differential \(du\). Using the product rule for differentiation on \(xe^x\):

\(\rm \frac{d}{dx}(xe^x) = (1 \cdot e^x + x \cdot e^x) = e^x(1+x)\)

Therefore, the differential \(du\) is:

\(du = \frac{d}{dx}(xe^x + 1) dx = e^x(x+1) dx\)

From this, we can express \((x+1) dx\) as:

\((x+1) dx = \frac{du}{e^x}\)

Also, from our substitution \(u = xe^x + 1\), we can write \(xe^x = u - 1\).

Now substitute these into the original integral:

\(\rm \int \frac{(x+1)}{x(xe^x + 1)}\ dx = \int \frac{1}{x(xe^x + 1)} \cdot (x+1) dx\)

Substituting \(u\) for \(xe^x + 1\) and \(\frac{du}{e^x}\) for \((x+1) dx\):

\(\rm \int \frac{1}{x \cdot u} \cdot \frac{du}{e^x} = \int \frac{1}{x e^x \cdot u} du\)

Substitute \(u-1\) for \(xe^x\):

\(\rm \int \frac{1}{(u-1)u} du\)

Step 2: Partial Fraction Decomposition

The integral is now in a form that can be solved using partial fractions. We need to decompose \(\frac{1}{u(u-1)}\):

\(\rm \frac{1}{u(u-1)} = \frac{A}{u} + \frac{B}{u-1}\)

Multiply both sides by \(u(u-1)\) to clear the denominators:

\(1 = A(u-1) + Bu\)

To find A and B, we can use convenient values for \(u\):

• Let \(u = 0\): \(1 = A(0-1) + B(0) \implies 1 = -A \implies A = -1\)
• Let \(u = 1\): \(1 = A(1-1) + B(1) \implies 1 = B \implies B = 1\)

So, the decomposition is:

\(\rm \frac{1}{u(u-1)} = \frac{-1}{u} + \frac{1}{u-1}\)

Step 3: Integrating the Expression

Now substitute the partial fractions back into the integral:

\(\rm \int \left( \frac{1}{u-1} - \frac{1}{u} \right) du\)

Integrate term by term:

\(= \rm \log|u-1| - \log|u| + C\)

Using the properties of logarithms (\(\rm \log a - \log b = \log(\frac{a}{b})\)):

\(= \rm \log\left|\frac{u-1}{u}\right| + C\)

Step 4: Substituting Back

Finally, substitute the original expression for \(u\) back into the result. Recall that \(u = xe^x + 1\), which means \(u-1 = xe^x\).

\(\rm \log\left|\frac{xe^x}{xe^x + 1}\right| + C\)

Assuming \(x\) and \(e^x\) are positive, \(xe^x+1\) is also positive, so we can remove the absolute value signs:

\(\rm \log\left(\frac{xe^x}{xe^x + 1}\right) + C\)

Conclusion

The result of the integration matches the fourth option provided.