Question

The value of \( \int \left(x + \frac{1}{x}\right)^3 dx \) is equal to:

• A. \( \frac{1}{4}\left(x+\frac{1}{x}\right)^4 + c \)
• B. \( \frac{x^4}{4} + \frac{3x^2}{2} + 3\log x - \frac{1}{2x^2} + c \)
• C. \( \frac{x^4}{4} + \frac{3x^2}{2} + 3\log x + \frac{1}{x^2} + c \)
• D. \( (x-x^{-1})^3 + c \)

Hint:

For integrals of polynomials or rational functions raised to a small integer power, it is often easiest to expand the expression algebraically first. This simplifies the problem to integrating a sum of power functions.

Answer 1

The Correct Option is B

Step 1: Expand the integrand using the binomial theorem \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \). \[ \left(x + \frac{1}{x}\right)^3 = x^3 + 3(x^2)\left(\frac{1}{x}\right) + 3(x)\left(\frac{1}{x^2}\right) + \left(\frac{1}{x}\right)^3 = x^3 + 3x + \frac{3}{x} + x^{-3} \] Step 2: Integrate the expanded expression term by term using the power rule. \[ \int \left(x^3 + 3x + 3x^{-1} + x^{-3}\right) dx = \frac{x^4}{4} + \frac{3x^2}{2} + 3\log|x| + \frac{x^{-2}}{-2} + c \] Step 3: Simplify the final expression. \[ \frac{x^4}{4} + \frac{3x^2}{2} + 3\log x - \frac{1}{2x^2} + c \]