JEE Main 2023 — Mathematics Calculus

Let $\displaystyle I={\int }_{-{\mathrm{log}}_{e}2}^{{\mathrm{log}}_{e}2}{e}^{x}[{\mathrm{log}}_{e}({e}^{x}+\sqrt{1+{e}^{2x}})]dx$I=∫−loge​2loge​2​ex[loge​(ex+1+e2x​)]dx

Let us substitute ${e}^{x}=t$ex=t

$$\Rightarrow {e}^{x}dx=dt$$⇒exdx=dt

Lower limit $={e}^{-{\mathrm{log}}_{e}(2)}=\frac{1}{2}$=e−loge​(2)=21​

Upper limit $={e}^{{\mathrm{log}}_{e}(2)}=2$=eloge​(2)=2

$$\Rightarrow I={\int }_{\frac{1}{2}}^{2}1\times {\mathrm{log}}_{e}[t+\sqrt{1+{t}^{2}}]dt$$⇒I=∫21​2​1×loge​[t+1+t2​]dt

Apply integration by-parts.

$$\int u(t)v(t)dt=u(t)\int v(t)dt-\int (u'(t)\int v(t)dt)dt$$∫u(t)v(t)dt=u(t)∫v(t)dt−∫(u′(t)∫v(t)dt)dt

Take $u(t)={\mathrm{log}}_{e}[t+\sqrt{1+{t}^{2}}],v(t)=1$u(t)=loge​[t+1+t2​],v(t)=1

$$\Rightarrow u'(t)=\frac{1}{t+\sqrt{1+{t}^{2}}}(1+\frac{2t}{2\sqrt{1+{t}^{2}}})=t$$⇒u′(t)=t+1+t2​1​(1+21+t2​2t​)=t

$$={[t\mathrm{ln}(\sqrt{{t}^{2}+1}+x)]}_{\frac{1}{2}}^{2}-{\int }_{1/2}^{2}\frac{t}{\sqrt{{t}^{2}+1}}dt$$=[tln(t2+1​+x)]21​2​−∫1/22​t2+1​t​dt

$$={[t\mathrm{ln}\sqrt{{t}^{2}+1}+t-\sqrt{{t}^{2}+1}]}_{\frac{1}{2}}^{2}$$=[tlnt2+1​+t−t2+1​]21​2​

$$=[2\mathrm{ln}(\sqrt{5}+2)-\sqrt{5}]-[\frac{1}{2}\mathrm{ln}(\sqrt{\frac{5}{2}}+\frac{1}{2})-\sqrt{\frac{5}{2}}]$$=[2ln(5​+2)−5​]−[21​ln(25​​+21​)−25​​]

$$=\mathrm{ln}(\frac{\sqrt{2}(2+\sqrt{5}{)}^{2}}{\sqrt{1+\sqrt{5}}})-\frac{\sqrt{5}}{2}$$=ln(1+5​​2​(2+5​)2​)−25​​

Hence this is the correct option.