Evaluate: `int(1+2x+3x^2+4x^3+)dx ,(<|x|<1)`

Explanation

To evaluate the integral

\[ \int (1 + 2x + 3x^2 + 4x^3 + \ldots) \, dx \quad \text{for } |x| < 1, \]

we start by recognizing that the series \(1 + 2x + 3x^2 + 4x^3 + \ldots\) can be expressed as a function.

### Step 1: Identify the series
The series can be identified as the power series for the function:

\[ \sum_{n=0}^{\infty} (n+1)x^n = \frac{1}{(1-x)^2} \quad \text{for } |x| < 1. \]

Thus, we have:

\[ 1 + 2x + 3x^2 + 4x^3 + \ldots = \frac{1}{(1-x)^2}. \]

### Step 2: Rewrite the integral
Now we can rewrite the integral as:

\[ \int (1 + 2x + 3x^2 + 4x^3 + \ldots) \, dx = \int \frac{1}{(1-x)^2} \, dx. \]

### Step 3: Integrate the function
To integrate \(\frac{1}{(1-x)^2}\), we can use the substitution \(u = 1 - x\), which gives us \(du = -dx\). Thus, the integral becomes:

\[ \int \frac{1}{u^2} (-du) = -\int u^{-2} \, du. \]

### Step 4: Solve the integral
The integral of \(u^{-2}\) is:

\[ -\int u^{-2} \, du = -\left(-\frac{1}{u}\right) + C = \frac{1}{u} + C. \]

Substituting back for \(u\):

\[ \frac{1}{1-x} + C. \]

### Final Answer
Thus, the evaluated integral is:

\[ \int (1 + 2x + 3x^2 + 4x^3 + \ldots) \, dx = \frac{1}{1-x} + C. \]