Verified Solution • 2025–26 Session moderate

QJEEM23Madd7d: JEE Main 2023 Mathematics - Class 12 JEEM23Madd7d Solution: ∫ 0 ∞ 6 e 3 x + 6 e 2 x + 11 e x + 6 d x =

∫0∞6e3x+6e2x+11ex+6dx=

Multiple Choice Options

  • A log⁡e(25681)\log _{e}\left(\frac{256}{81}\right)
  • B log⁡e(6427)\log _{e}\left(\frac{64}{27}\right)
  • C log⁡e(3227)\log _{e}\left(\frac{32}{27}\right)
  • D log⁡e(51281)\log _{e}\left(\frac{512}{81}\right)

Step-by-Step Verified Solution

amp;l=∫0∞6(ex+1)(ex+2)(ex+3)dxamp;=6∫0∞(12ex+1+−1ex+2+12ex+3)dxamp;=3∫0∞e−x1+e−xdx−6∫0∞e−xdx1+2e−x+3∫0∞e−x1+3e−xdxamp;=3[−ln⁡(1+e−x)]0∞+612[ln⁡(1+2e−x)]0∞−33[ln⁡(1+3e−x)]0∞amp;=3ln⁡2−3ln⁡3+ln⁡4amp;=3ln⁡23+ln⁡4amp;=ln⁡3227\begin{aligned}& \mathrm{l}=\int_0^{\infty} \frac{6}{\left(\mathrm{e}^{\mathrm{x}}+1\right)\left(\mathrm{e}^{\mathrm{x}}+2\right)\left(\mathrm{e}^{\mathrm{x}}+3\right)} \mathrm{dx} \\\\& =6 \int_0^{\infty}\left(\frac{\frac{1}{2}}{\mathrm{e}^{\mathrm{x}}+1}+\frac{-1}{\mathrm{e}^{\mathrm{x}}+2}+\frac{\frac{1}{2}}{\mathrm{e}^{\mathrm{x}}+3}\right) \mathrm{dx} \\\\& =3 \int_0^{\infty} \frac{\mathrm{e}^{-\mathrm{x}}}{1+\mathrm{e}^{-\mathrm{x}}} \mathrm{dx}-6 \int_0^{\infty} \frac{\mathrm{e}^{-\mathrm{x}} \mathrm{dx}}{1+2 \mathrm{e}^{-\mathrm{x}}}+3 \int_0^{\infty} \frac{\mathrm{e}^{-\mathrm{x}}}{1+3 \mathrm{e}^{-\mathrm{x}}} \mathrm{dx} \\\\& =3\left[-\ln \left(1+\mathrm{e}^{-\mathrm{x}}\right)\right]_0^{\infty}+6 \frac{1}{2}\left[\ln \left(1+2 \mathrm{e}^{-\mathrm{x}}\right)\right]_0^{\infty} -\frac{3}{3}\left[\ln \left(1+3 \mathrm{e}^{-\mathrm{x}}\right)\right]_0^{\infty} \\\\& =3 \ln 2-3 \ln 3+\ln 4 \\\\& =3 \ln \frac{2}{3}+\ln 4 \\\\& =\ln \frac{32}{27}\end{aligned}

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