_0^6e^3x+6e^2x+11e^x+6,dx equals:

Question

$\displaystyle\int_0^\infty\frac{6}{e^{3x}+6e^{2x}+11e^x+6}\,dx$ equals:

Accepted Answer

Factor the denominator: $e^{3x}+6e^{2x}+11e^x+6=(e^x+1)(e^x+2)(e^x+3)$. Partial fractions:

$$\frac6{(e^x+1)(e^x+2)(e^x+3)} = \frac{3}{e^x+1}-\frac{6}{e^x+2}+\frac{3}{e^x+3}$$

Multiplying through by $e^{-x}$ and integrating each term (each of the form $\int\frac{e^{-x}}{1+ke^{-x}}dx=-\frac1k\ln(1+ke^{-x})$):

$$I = \left[3\cdot\!\big(-\ln(1+e^{-x})\big)-6\cdot\left(-\frac12\ln(1+2e^{-x})\right)+3\cdot\left(-\frac13\ln(1+3e^{-x})\right)\right]_0^\infty$$

$$= 3\ln2-3\ln3+\ln4 = 3\ln\frac23+\ln4 = \log_e\frac{32}{27}$$

Solutions · 1