If `f(x)=lim_(n->oo)[2x+4x^3+6x^5++2n x^(2n-1)]` `(0ltxlt1)` then `int f(x)dx` is equal to:

Options

A. `-sqrt(1-x^(2))`
B. `(1)/(sqrt(1-x^(2)))`
C. `(1)/((x^(2)-1))`
D. `(1)/(1-x^(2))`

Answer

To solve the problem, we need to evaluate the function \( f(x) \) defined as:

\[ f(x) = \lim_{n \to \infty} \left( 2x + 4x^2 + 6x^3 + \ldots + 2n x^{2n-1} \right) \]

for \( 0 < x < 1 \), and then find the integral \( \int f(x) \, dx \).

### Step 1: Rewrite the function as a summation
The expression inside the limit can be rewritten as a summation:

\[ f(x) = \lim_{n \to \infty} \sum_{k=1}^{n} 2k x^{2k-1} \]

### Step 2: Factor out the constant
We can factor out the constant \( 2 \):

\[ f(x) = 2 \lim_{n \to \infty} \sum_{k=1}^{n} k x^{2k-1} \]

### Step 3: Use the formula for the sum of a series
The sum \( \sum_{k=1}^{n} k x^{k} \) can be evaluated using the formula:

\[ \sum_{k=1}^{n} k x^{k} = x \frac{d}{dx} \left( \sum_{k=0}^{n} x^{k} \right) \]

The sum \( \sum_{k=0}^{n} x^{k} \) is a geometric series:

\[ \sum_{k=0}^{n} x^{k} = \frac{1 - x^{n+1}}{1 - x} \]

### Step 4: Differentiate the geometric series
Differentiating with respect to \( x \):

\[ \frac{d}{dx} \left( \frac{1 - x^{n+1}}{1 - x} \right) = \frac{(1 - x)(-(n+1)x^n) - (1 - x^{n+1})(-1)}{(1 - x)^2} \]

This simplifies to:

\[ \frac{(n+1)x^n - (n+1)x^{n+1} + 1 - x^{n+1}}{(1 - x)^2} \]

### Step 5: Evaluate the limit as \( n \to \infty \)
As \( n \to \infty \) and since \( 0 < x < 1 \), the terms involving \( x^n \) vanish:

\[ \lim_{n \to \infty} \sum_{k=1}^{n} k x^{k} = \frac{x}{(1 - x)^2} \]

Thus, we have:

\[ f(x) = 2 \cdot \frac{x}{(1 - x)^2} \]

### Step 6: Find the integral \( \int f(x) \, dx \)
Now we need to find:

\[ \int f(x) \, dx = \int \frac{2x}{(1 - x)^2} \, dx \]

### Step 7: Use substitution
Let \( u = 1 - x \), then \( du = -dx \) and \( x = 1 - u \):

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

### Step 8: Integrate
This gives:

\[ -2 \left( \ln |u| + \frac{1}{u} \right) + C = -2 \left( \ln |1 - x| + \frac{1}{1 - x} \right) + C \]

### Final Result
Thus, the integral \( \int f(x) \, dx \) is:

\[ -2 \ln(1 - x) + \frac{2}{1 - x} + C \]