Question

The integral \( \int \left( \frac{x}{2} - \frac{x^2}{3} + \frac{x^3}{4} - \cdots \right) \, dx \) is equal to

• A. \( \frac{1}{1+x} + \text{Constant} \)
• B. \( \frac{1}{1+x^2} + \text{Constant} \)
• C. \( - \frac{1}{1-x} + \text{Constant} \)
• D. \( - \frac{1}{1-x^2} + \text{Constant} \)

Hint:

When dealing with series integrals, identify common series expansions (such as arctangent) to simplify and identify the answer quickly.

Answer 1

The Correct Option is B

The given integral is: \[ \int \left( \frac{x}{2} - \frac{x^2}{3} + \frac{x^3}{4} - \cdots \right) dx \] This is a series of terms, where each term is of the form \( \frac{x^n}{n+1} \). The general term for this series is \( \frac{x^n}{n+1} \), and we can integrate each term separately. For the term \( \frac{x^n}{n+1} \), the integral is: \[ \int \frac{x^n}{n+1} dx = \frac{x^{n+1}}{(n+1)(n+2)} + C \] Let’s look at the first few terms of the series and integrate each one: - For the first term \( \frac{x}{2} \), the integral is: \[ \int \frac{x}{2} dx = \frac{x^2}{4} + C \] - For the second term \( - \frac{x^2}{3} \), the integral is: \[ \int - \frac{x^2}{3} dx = - \frac{x^3}{9} + C \] - For the third term \( \frac{x^3}{4} \), the integral is: \[ \int \frac{x^3}{4} dx = \frac{x^4}{16} + C \] Thus, the integral of the entire series is: \[ \int \left( \frac{x}{2} - \frac{x^2}{3} + \frac{x^3}{4} - \cdots \right) dx = \frac{x^2}{4} - \frac{x^3}{9} + \frac{x^4}{16} - \cdots \] Now, let's compare this result with the given options. Upon closer inspection, it matches the form of the inverse tangent function expansion. The series expansion for the arctangent function \( \arctan(x) \) is: \[ \arctan(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} \] This is the series form of the integrals of the terms of the given series. Therefore, the correct answer corresponds to Option (B). Thus, the value of \( x \) is \( \boxed{\text{Option (B)}} \).