If $y=\int(log_{e}x)^{2}dx$ satisfies the point (1, 2), then :

Question

Accepted Answer

To determine why Option 3 is the correct answer, we need to analyze the integral $ y = \int (\log_e x)^2 \, dx $ and verify how it satisfies the condition given by the point (1, 2).

### Step-by-Step Explanation:

1. **Integration of the Function:**
   The integral $ y = \int (\log_e x)^2 \, dx $ can be solved using integration by parts. Let $ u = (\log_e x)^2 $ and $ dv = dx $. Then, $ du = 2 \log_e x \cdot \frac{1}{x} \, dx $ and $ v = x $.

Using integration by parts:
   $$
   \int u \, dv = uv - \int v \, du
   $$
   $$
   y = x(\log_e x)^2 - \int x \cdot 2 \log_e x \cdot \frac{1}{x} \, dx
   $$
   $$
   y = x(\log_e x)^2 - 2 \int \log_e x \, dx
   $$

2. **Integrating $\int \log_e x \, dx$:**
   Again, use integration by parts with $ u = \log_e x $ and $ dv = dx $. Then, $ du = \frac{1}{x} \, dx $ and $ v = x $.
   $$
   \int \log_e x \, dx = x \log_e x - \int x \cdot \frac{1}{x} \, dx = x \log_e x - x + C
   $$

3. **Substitute Back:**
   $$
   y = x(\log_e x)^2 - 2(x \log_e x - x)
   $$
   $$
   y = x(\log_e x)^2 - 2x \log_e x + 2x
   $$

4. **Simplify and Apply the Condition:**
   We know that $ y $ must satisfy the point (1, 2). Substitute $ x = 1 $ and $ y = 2 $:
   $$
   2 = 1(\log_e 1)^2 - 2 \cdot 1 \cdot \log_e 1 + 2 \cdot 1
   $$
   Since $\log_e 1 = 0$, this simplifies to:
   $$
   2 = 0 - 0 + 2
   $$
   This confirms the condition is satisfied.

5. **Verify Option 3:**
   Option 3 states:
   $$
   \frac{y}{x} - 2 = \log_e x (\log_e x - 2)
   $$
   Substitute $ y = x(\log_e x)^2 - 2x \log_e x + 2x $ into the left side:
   $$
   \frac{x(\log_e x)^2 - 2x \log_e x + 2x}{x} - 2 = (\log_e x)^2 - 2 \log_e x + 2 - 2
   $$
   $$
   = (\log_e x)^2 - 2 \log_e x = \log_e x (\log_e x - 2)
   $$
   This matches the right side of Option 3, confirming its correctness.

### Explanation of Incorrect Options:

- **Option 1:** $\frac{y}{x} - 2 = \log_e x (\log_e x + 2)$
  - The expression $(\log_e x)^2 - 2 \log_e x$ does not simplify to $\log_e x (\log_e x + 2)$.

- **Option 2:** $\frac{y}{x} + 2 = \log_e x (\log_e x + 2)$
  - Adding 2 to the left side does not match the derived expression.

- **Option 4:** $\frac{y}{x} + 2 = \log_e x (\log_e x - 2)$
  - Adding 2 to the left side results in a mismatch with the derived expression.

Thus, Option 3 is the only one that correctly represents the relationship derived from the integral and the given point.