If `f(x)=1-x+x^(2)-x^(3)+x^(4)+….oo` then
`int_(0)^(x)f(x)dx=`

Options

A. log x
B. `log(1-x)`
C. `log(1+x)`
D. `log(1+x^(2))`

Answer

To solve the problem, we need to evaluate the integral \( \int_{0}^{x} f(t) \, dt \), where \( f(x) = 1 - x + x^2 - x^3 + x^4 - \ldots \).

### Step 1: Identify the function \( f(x) \)

The function \( f(x) \) is an infinite series:
\[ f(x) = 1 - x + x^2 - x^3 + x^4 - \ldots \]
This series can be recognized as a geometric series with the first term \( a = 1 \) and the common ratio \( r = -x \).

### Step 2: Sum the infinite series

The sum of an infinite geometric series can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
provided that \( |r| < 1 \). In our case:
\[ f(x) = \frac{1}{1 - (-x)} = \frac{1}{1 + x} \quad \text{for } |x| < 1 \]

### Step 3: Set up the integral

Now we can set up the integral:
\[ \int_{0}^{x} f(t) \, dt = \int_{0}^{x} \frac{1}{1 + t} \, dt \]

### Step 4: Evaluate the integral

The integral of \( \frac{1}{1 + t} \) is:
\[ \int \frac{1}{1 + t} \, dt = \ln(1 + t) + C \]
Thus, we evaluate the definite integral:
\[ \int_{0}^{x} \frac{1}{1 + t} \, dt = \left[ \ln(1 + t) \right]_{0}^{x} = \ln(1 + x) - \ln(1 + 0) = \ln(1 + x) - \ln(1) = \ln(1 + x) \]

### Final Answer

Therefore, the result of the integral is:
\[ \int_{0}^{x} f(t) \, dt = \ln(1 + x) \]

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