JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

\int_{-\log_e 2}^{\log_e 2} e^x \left( \log_e \left( e^x + \sqrt{1 + e^{2x}} \right) \right) dx

Choose the correct answer:

Explanation

Step 1: Substitution

Maana $t = e^x$ , toh $dt = e^x dx$ .

Ab limits badalte hain:

Lower limit: Jab $x = -\log_e 2 \implies t = e^{\log_e 2^{-1}} = \frac{1}{2}$
Upper limit: Jab $x = \log_e 2 \implies t = 2$

Ab integral ye ban jayega:

I = \int_{1/2}^{2} \log_e (t + \sqrt{1+t^2}) dt

Step 2: Integration by Parts

Hum $\int u \cdot v \, dt$ ka formula lagayenge, jahan $u = \log_e(t + \sqrt{1+t^2})$ aur $v = 1$ hai.

Hame pata hai ki $\frac{d}{dt}[\log_e(t + \sqrt{1+t^2})] = \frac{1}{\sqrt{1+t^2}}$ .

I = \left[ t \log_e(t + \sqrt{1+t^2}) \right]_{1/2}^{2} - \int_{1/2}^{2} \frac{t}{\sqrt{1+t^2}} dt

Step 3: Limits aur Integration apply karna

Pehle part mein limits rakhne par:

= \left( 2 \log_e(2+\sqrt{5}) - \frac{1}{2} \log_e\left(\frac{1}{2} + \sqrt{1+\frac{1}{4}}\right) \right)

= 2 \log_e(2+\sqrt{5}) - \frac{1}{2} \log_e\left(\frac{1+\sqrt{5}}{2}\right)

Doosre part ka integral:

\int \frac{t}{\sqrt{1+t^2}} dt = \sqrt{1+t^2}

Limits lagane par:

- \left[ \sqrt{1+t^2} \right]_{1/2}^{2} = - \left( \sqrt{5} - \sqrt{\frac{5}{4}} \right) = - \left( \sqrt{5} - \frac{\sqrt{5}}{2} \right) = -\frac{\sqrt{5}}{2}

Step 4: Final Simplification

Sabko ek saath likhne par:

I = \log_e(2+\sqrt{5})^2 - \log_e\left(\frac{1+\sqrt{5}}{2}\right)^{1/2} - \frac{\sqrt{5}}{2}

I = \log_e \left( \frac{(2+\sqrt{5})^2}{\sqrt{\frac{1+\sqrt{5}}{2}}} \right) - \frac{\sqrt{5}}{2}

I = \log_e \left( \frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}} \right) - \frac{\sqrt{5}}{2}

Choose the correct answer:

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