JEE Main 2023 — Mathematics Calculus

Let

$$I={\int }_{0}^{\infty }(\frac{6}{{e}^{3x}+6{e}^{2x}+11{e}^{x}+6})dx$$I=∫0∞​(e3x+6e2x+11ex+66​)dx

$$\Rightarrow I={\int }_{0}^{\infty }\frac{6}{({e}^{x}+1)({e}^{x}+2)({e}^{x}+3)}dx$$⇒I=∫0∞​(ex+1)(ex+2)(ex+3)6​dx

Using partial fraction, we can write

$$\Rightarrow I=6{\int }_{0}^{\infty }[\frac{1}{2({e}^{x}+1)}-\frac{1}{({e}^{x}+2)}+\frac{1}{2({e}^{x}+3)}]dx$$⇒I=6∫0∞​[2(ex+1)1​−(ex+2)1​+2(ex+3)1​]dx

$$\Rightarrow I=6{\int }_{0}^{\infty }[\frac{1}{2({e}^{-x}+1)}-\frac{1}{(1+2{e}^{-x})}+\frac{1}{2(1+3{e}^{-x})}]{e}^{-x}dx$$⇒I=6∫0∞​[2(e−x+1)1​−(1+2e−x)1​+2(1+3e−x)1​]e−xdx

$$\Rightarrow I=6{[-\frac{1}{2}{\mathrm{log}}_{e}({e}^{-x}+1)+\frac{1}{2}{\mathrm{log}}_{e}(2{e}^{-x}+1)-\frac{1}{6}{\mathrm{log}}_{e}(3{e}^{-x}+1)]}_{0}^{\infty }$$⇒I=6[−21​loge​(e−x+1)+21​loge​(2e−x+1)−61​loge​(3e−x+1)]0∞​

$$\Rightarrow I=6[0-{-\frac{1}{2}{\mathrm{log}}_{e}2+\frac{1}{2}{\mathrm{log}}_{e}3-\frac{1}{6}{\mathrm{log}}_{e}4}]$$⇒I=6[0−−21​loge​2+21​loge​3−61​loge​4]

$$\Rightarrow I=3{\mathrm{log}}_{e}2-3{\mathrm{log}}_{e}3+{\mathrm{log}}_{e}4$$⇒I=3loge​2−3loge​3+loge​4

$$\Rightarrow I=5{\mathrm{log}}_{e}2-3{\mathrm{log}}_{e}3$$⇒I=5loge​2−3loge​3

$$\Rightarrow I={\mathrm{log}}_{e}(\frac{32}{27})$$⇒I=loge​(2732​)