Question

\(\int^{1}_{0}\frac{2e^x}{1+e^{2x}}dx=\)

• A. \(4(\tan^{-1}2-\pi)\)
• B. \(2(\tan^{-1}e-\frac{\pi}{2})\)
• C. \(2(\tan^{-1}e+\frac{\pi}{4})\)
• D. \(2(\tan^{-1}e-\frac{\pi}{4})\)
• E. \(2(\tan^{-1}2+\pi)\)

Answer 1

The Correct Option is D

Step 1: Simplify the integrand: \[ \frac{2e^x}{1 + e^{2x}} = \frac{2e^x}{1 + (e^x)^2} \]

Step 2: Make the substitution: \[ u = e^x \quad \Rightarrow \quad du = e^x dx \] When \( x = 0 \), \( u = 1 \) When \( x = 1 \), \( u = e \)

Step 3: Transform the integral: \[ \int_{1}^{e} \frac{2}{1 + u^2} du \]

Step 4: Recognize the standard integral form: \[ 2 \int_{1}^{e} \frac{1}{1 + u^2} du = 2 \left[ \tan^{-1} u \right]_{1}^{e} \]

Step 5: Evaluate the integral: \[ 2 \left( \tan^{-1} e - \tan^{-1} 1 \right) = 2 \left( \tan^{-1} e - \frac{\pi}{4} \right) \]

Step 6: Compare with given options: The expression \( 2 \left( \tan^{-1} e - \frac{\pi}{4} \right) \) can be rewritten as: \[ 2 \tan^{-1} e - \frac{\pi}{2} \] which matches option (D).

Conclusion: The correct answer is \(\boxed{D}\) \(\left( 2(\tan^{-1} e - \frac{\pi}{2}) \right)\).

Answer 2

Let the given integral be

\[ I = \int_0^1 \frac{2e^x}{1 + e^{2x}} dx \]

Let \( u = e^x \). Then \( du = e^x dx \). When \( x = 0 \), \( u = 1 \). When \( x = 1 \), \( u = e \).

Then the integral becomes

\[ I = \int_1^e \frac{2}{1 + u^2} du = 2 \int_1^e \frac{1}{1 + u^2} du \]

We know that \( \int \frac{1}{1+u^2} du = \arctan(u) + C \). Therefore,

\[ I = 2 [\arctan(u)]_1^e = 2 (\arctan(e) - \arctan(1)) = 2 (\arctan(e) - \frac{\pi}{4}) \]

We can approximate this value:

\[ \arctan(e) \approx \arctan(2.718) \approx 1.217 \text{ radians.} \] \[ \text{Then } I \approx 2(1.217 - \frac{\pi}{4}) \approx 2(1.217 - 0.785) \approx 2(0.432) \approx 0.864. \]

The exact value is \( 2(\arctan(e) - \frac{\pi}{4}) \).

Final Answer: The final answer is \( {2(\arctan(e)-\frac{\pi}{4})} \).