$\int\frac{(x+1)(x+log~x)^{2}}{x}dx=$

Question

Accepted Answer

To solve the integral $\int\frac{(x+1)(x+\log x)^{2}}{x}dx$, we can use substitution to simplify the expression. Let's go through the steps to understand why Option 2 is the correct answer.

### Explanation:

1. **Substitution:**
   Let $ u = x + \log x $. Then, the derivative $ du $ is given by:
   $$
   du = \left(1 + \frac{1}{x}\right) dx = \frac{x+1}{x} dx
   $$
   This implies:
   $$
   dx = \frac{x}{x+1} du
   $$

2. **Rewriting the Integral:**
   Substitute $ u = x + \log x $ into the integral:
   $$
   \int \frac{(x+1)(x+\log x)^{2}}{x} dx = \int u^2 \cdot \frac{x+1}{x} dx
   $$
   Using the expression for $ dx $ from the substitution:
   $$
   = \int u^2 \cdot \frac{x+1}{x} \cdot \frac{x}{x+1} du = \int u^2 du
   $$

3. **Integrating:**
   The integral $\int u^2 du$ is straightforward:
   $$
   \int u^2 du = \frac{u^3}{3} + C
   $$

4. **Back-Substitution:**
   Substitute back $ u = x + \log x $:
   $$
   \frac{(x + \log x)^3}{3} + C
   $$

Thus, the integral evaluates to:
$$
\frac{1}{3}(x + \log x)^3 + C
$$

### Justification for Option 2:

- **Option 2**: $\frac{1}{3}(x+\log x)^{3}+c$ matches exactly with our derived result, confirming it as the correct answer.

### Explanation for Incorrect Options:

- **Option 1**: $\frac{1}{3}(x+\log x)^{2}+c$ is incorrect because the power of the expression inside the integral is not reduced correctly. The integration of $u^2$ results in $\frac{u^3}{3}$, not $\frac{u^2}{3}$.

- **Option 3**: $\frac{2}{3}(x+\log x)^{2}+c$ is incorrect for similar reasons as Option 1. Additionally, the coefficient $\frac{2}{3}$ does not appear in the integration of $u^2$.

- **Option 4**: $\frac{2}{3}(x+\log x)^{3}+c$ is incorrect because the coefficient $\frac{2}{3}$ is not derived from the integration process. The correct coefficient is $\frac{1}{3}$.

In conclusion, Option 2 is the correct answer because it accurately represents the result of the integration process, considering the substitution and integration steps.