If $y(x)=\int(1+2x+3x^{2}+....)dx,$ $|x|

Question

If $y(x)=\int(1+2x+3x^{2}+....)dx,$ $|x|<1$ and $y(0)=0$, then $y(x)$ is equal to:

Accepted Answer

To solve the problem and justify why Option 2 is the correct answer, we need to evaluate the integral given in the problem and verify that it matches the expression provided in Option 2.

### Explanation:

1. **Understanding the Series:**
   The series given inside the integral is $1 + 2x + 3x^2 + \ldots$. This is a power series where the coefficient of $x^n$ is $n+1$. This series can be recognized as the derivative of a geometric series. Specifically, the series $1 + x + x^2 + \ldots$ is the geometric series with the sum $\frac{1}{1-x}$ for $|x| < 1$.

2. **Differentiating the Geometric Series:**
   The derivative of $\frac{1}{1-x}$ with respect to $x$ is:
   $$
   \frac{d}{dx}\left(\frac{1}{1-x}\right) = \frac{1}{(1-x)^2}
   $$
   However, the series given is $1 + 2x + 3x^2 + \ldots$, which is the derivative of the series $1 + x + x^2 + \ldots$ multiplied by $x$. Therefore, the series can be expressed as:
   $$
   \sum_{n=0}^{\infty} (n+1)x^n = \frac{1}{(1-x)^2}
   $$

3. **Integrating the Series:**
   We need to integrate the series term by term:
   $$
   \int (1 + 2x + 3x^2 + \ldots) \, dx = \int \frac{1}{(1-x)^2} \, dx
   $$
   The integral of $\frac{1}{(1-x)^2}$ is:
   $$
   \int \frac{1}{(1-x)^2} \, dx = -\frac{1}{1-x} + C
   $$

4. **Applying the Initial Condition:**
   We are given that $y(0) = 0$. Substituting $x = 0$ into the integrated function:
   $$
   -\frac{1}{1-0} + C = 0 \implies -1 + C = 0 \implies C = 1
   $$
   Therefore, the function becomes:
   $$
   y(x) = -\frac{1}{1-x} + 1 = \frac{x}{1-x}
   $$

5. **Conclusion:**
   The expression $\frac{x}{1-x}$ matches Option 2, confirming it as the correct answer.

### Why Other Options are Incorrect:

- **Option 1: $\frac{1}{1-x}$**
  - This is the sum of the geometric series $1 + x + x^2 + \ldots$, not the integral of the given series.

- **Option 3: $\frac{1}{(1-x)^2}$**
  - This represents the derivative of the geometric series, not the integral.

- **Option 4: $\frac{x}{(1-x)^2}$**
  - This would be the result of differentiating $\frac{x}{1-x}$, not integrating the given series.

Thus, the correct answer is indeed Option 2: $\frac{x}{1-x}$.