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QJEEM23Ma8cd2: JEE Main 2023 Mathematics - Class 12 JEEM23Ma8cd2 Solution: The value of the integral ∫ − log e ⁡ 2 log e ⁡ 2 e x ( log e ⁡ …

The value of the integral ∫−loge⁡2loge⁡2ex(loge⁡(ex+1+e2x))dx is equal to :

Multiple Choice Options

  • A log⁡e((2+5)21+5)+52\log _{e}\left(\frac{(2+\sqrt{5})^{2}}{\sqrt{1+\sqrt{5}}}\right)+\frac{\sqrt{5}}{2}
  • B log⁡e(2(2+5)21+5)−52\log _{e}\left(\frac{\sqrt{2}(2+\sqrt{5})^{2}}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}
  • C log⁡e(2(2+5)1+5)−52\log _{e}\left(\frac{2(2+\sqrt{5})}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}
  • D log⁡e(2(3−5)21+5)+52\log _{e}\left(\frac{\sqrt{2}(3-\sqrt{5})^{2}}{\sqrt{1+\sqrt{5}}}\right)+\frac{\sqrt{5}}{2}

Step-by-Step Verified Solution

∫−loge⁡2loge⁡2ex(loge⁡(ex+1+e2x))dx

Let ex=t⇒exdx=dte^x=t \Rightarrow e^x d x=d t

When, x→−log⁡e2x \rightarrow-\log _e 2 , then t→12t \rightarrow \frac{1}{2}

When, x→log⁡e2x \rightarrow \log _e 2 , then t→2t \rightarrow 2

I=∫122[loge⁡(t+1+t2)]dt ...........(i)

On applying integration by part method in Eq. (i), we get

amp;I=[tlog⁡e(t+1+t2)]1/22−∫1/22tt+1+t2(1+2t21+t2)dtamp;=2log⁡e(2+5)−12log⁡e(1+52)−∫1/22t1+t2dt\begin{aligned}& I=\left[t \log _e\left(t+\sqrt{1+t^2}\right)\right]_{1 / 2}^2-\int_{1 / 2}^2 \frac{t}{t+\sqrt{1+t^2}}\left(1+\frac{2 t}{2 \sqrt{1+t^2}}\right) d t \\\\& =2 \log _e(2+\sqrt{5})-\frac{1}{2} \log _e\left(\frac{1+\sqrt{5}}{2}\right)-\int_{1 / 2}^2 \frac{t}{\sqrt{1+t^2}} d t\end{aligned}

=log⁡e((2+5)2(1+52)1/2)−12∫1/222t1+t2dt=\log _e\left(\frac{(2+\sqrt{5})^2}{\left(\frac{1+\sqrt{5}}{2}\right)^{1 / 2}}\right)-\frac{1}{2} \int_{1 / 2}^2 \frac{2 t}{\sqrt{1+t^2}} d t .............(ii)

Let I1=∫1/222t1+t2dt\quad I_1=\int_{1 / 2}^2 \frac{2 t}{\sqrt{1+t^2}} d t

Let 1+t2=w1+t^2=w

2tdt=dw2 t d t=d w

When, t→12t \rightarrow \frac{1}{2} , then w=54w=\frac{5}{4}

When, t→2t \rightarrow 2 , then w=5w=5

I1amp;=∫5/451wdwamp;=[2w]5/45amp;=2[5−52]=5\begin{aligned}I_1 & =\int_{5 / 4}^5 \frac{1}{\sqrt{w}} d w \\\\& =[2 \sqrt{w}]_{5 / 4}^5 \\\\& =2\left[\sqrt{5}-\frac{\sqrt{5}}{2}\right]=\sqrt{5}\end{aligned}

On substitute value of I1I_1 in Eq. (ii), we get

I=log⁡e(2(2+5)21+5)−52I=\log _e\left(\frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}

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