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JEE Main Mathematics question on Indefinite Integration

∫(1+x−1x)ex+1xdx\int \left(1 + x - \frac{1}{x}\right) e^{x + \frac{1}{x}} dx is equal to
  1. A.
    (x−1)ex+1x+c(x-1) e^{x + \frac{1}{x}} + c
  2. B.
    xex+1x+cx e^{x + \frac{1}{x}} + c
  3. C.
    (x+1)ex+1x+c(x+1) e^{x + \frac{1}{x}} + c
  4. D.
    −xex+1x+c-x e^{x + \frac{1}{x}} + c
Show answer and solution

Correct option: B

Solution

Split the integrand as e^(x+1/x) + (x - 1/x)e^(x+1/x). Recognize the pattern ∫[f(x) + f'(x)]ex dx = f(x)ex + c, but here with exponent x+1/x. Let f(x) = x, then f'(x) = 1, and the derivative of (x+1/x) is (1 - 1/x²). Notice x·(1 - 1/x²) = x - 1/x, so d/dx[x·e^(x+1/x)] = e^(x+1/x) + x(1-1/x²)e^(x+1/x) = (1 + x - 1/x)e^(x+1/x), which exactly matches the integrand. Therefore the answer is x·e^(x+1/x) + c.

Previous-year question, JEE Main.

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