Question

The integral \(∫^∞_0  \frac{6}{(e^{3x} + 6e^{2x} + 11e{x} + 6)}dx\)

• A. ln \(32\)
• B. ln \(27\)
• C. ln \(\frac{32}{27}\)
• D. ln \(\frac{27}{32}\)

Answer 1

The Correct Option is C

Using partial fractions, we can write \[ \frac{6}{(e^x + 1)(e^x + 2)(e^x + 3)} = \frac{A}{e^x + 1} + \frac{B}{e^x + 2} + \frac{C}{e^x + 3}. \] Solving for \(A, B, C\) gives \[ A = \tfrac{1}{2}, \quad B = -1, \quad C = \tfrac{1}{2}. \] Hence \[ \frac{6}{(e^x + 1)(e^x + 2)(e^x + 3)} = \frac{1/2}{e^x + 1} - \frac{1}{e^x + 2} + \frac{1/2}{e^x + 3}. \] Therefore the integral becomes \[ I = \int_{0}^{\infty} \frac{6}{(e^x + 1)(e^x + 2)(e^x + 3)} \, dx = \int_{0}^{\infty} \left(\frac{1/2}{e^x + 1} - \frac{1}{e^x + 2} + \frac{1/2}{e^x + 3}\right) \, dx. \] Each term can be integrated via the substitution \(u = e^x\), \(du = e^x \, dx\). Carrying out these integrals and evaluating from \(x=0\) to \(x \to \infty\) yields \[ I = \ln\left(\tfrac{32}{27}\right). \] \[ \boxed{ \int_{0}^{\infty} \frac{6}{(e^x + 1)(e^x + 2)(e^x + 3)} \, dx = \ln\left(\tfrac{32}{27}\right). } \]