The value of the integral 2^- 2 e^x ( ( e^x + 1 + e^2x)) dx is equal to:

Given: I = 2^- 2 e^x ( ( e^x + 1 + e^2x)) dx Find: The value of the integral and the correct option. Use the substitution shown in the solution: t = e^x Then e^x dx = dt The limits change as follows: - when x = 2, t = 2 - when x = - 2, t = 12 So the integral becomes I = 2^ 12 (t + 1+t^2) dt Now, applying integration by parts as stated in the solution, we get I = [ ( t + t^2 + 1) ] 12^2 Evaluating the expression gives ( 2 + 5 5) / 2 Therefore, the correct option is D. When solving integrals involving logarithms and square roots, consider substitution and integration by parts.

The value of the integral 2^- 2 e^x ( ( e^x + 1 + e^2x)) dx is eq… — Solution | JEE Mathematics

MCQMediumJEE 2023Indefinite Integrals

The value of the integral $\int_{\log 2}^{-\log 2} e^x \left( \log \left( e^x + \sqrt{1 + e^{2x}} \right) \right) dx$ is equal to:

A
$\log \left( \frac{2 + \sqrt{5}}{\sqrt{5}} \right)$
B
$\log \left( \frac{2 + \sqrt{5}}{\sqrt{5}} \right) / 2$
C
$\log \left( \frac{2\sqrt{5}}{\sqrt{5}} \right)$
D
$\log \left( \frac{2 + \sqrt{5}}{\sqrt{5}} \right) / 2$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

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

Find: The value of the integral and the correct option.

Use the substitution shown in the solution:

t = e^x

Then

e^x \, dx = dt

The limits change as follows:

when $x = \log 2$ , $t = 2$
when $x = -\log 2$ , $t = \frac{1}{2}$

So the integral becomes

I = \int_{2}^{\frac{1}{2}} \log\left(t + \sqrt{1+t^2}\right) \, dt

Now, applying integration by parts as stated in the solution, we get

I = \left[ \ln \left( t + \sqrt{t^2 + 1} \right) \right]_{\frac{1}{2}}^{2}

Evaluating the expression gives

\Rightarrow \log \left( \frac{2 + \sqrt{5}}{\sqrt{5}} \right) / 2

Therefore, the correct option is D.

Extracted Hint

When solving integrals involving logarithms and square roots, consider substitution and integration by parts.

Common mistakes

Changing the limits incorrectly after substituting $t=e^x$ . Since the original limits are $x=\log 2$ to $x=-\log 2$ , the new limits become $t=2$ to $t=\frac{1}{2}$ , not the other way around.
Forgetting that $e^x dx = dt$ under the substitution $t=e^x$ . If this differential change is missed, the transformed integral is written incorrectly.
Confusing $\log\left(t+\sqrt{1+t^2}\right)$ with an elementary algebraic expression and trying to simplify it directly. The intended method in the solution is substitution followed by integration by parts.

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Answer

Step-by-step solution

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