The value of the integral int limits_{-log _{ | vedclass

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(D) Let $I = \int \limits_{-\ln 2}^{\ln 2} e^x \ln \left(e^x+\sqrt{1+e^{2 x}}\right) d x$.Substitute $e^x = t$,then $e^x dx = dt$. When $x = -\ln 2$,$

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$\log _e\left(\frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}$

Vedclass · Answer

(D) Let $I = \int \limits_{-\ln 2}^{\ln 2} e^x \ln \left(e^x+\sqrt{1+e^{2 x}}\right) d x$.Substitute $e^x = t$,then $e^x dx = dt$. When $x = -\ln 2$,$t = 1/2$. When $x = \ln 2$,$t = 2$.$I = \int \limits_{1/2}^{2} \ln \left(t+\sqrt{1+t^2}\right) dt$.Using integration by parts,$\int u dv = uv - \int v du$. Let $u = \ln(t+\sqrt{1+t^2})$ and $dv = dt$.Then $du = \frac{1}{t+\sqrt{1+t^2}} \left(1 + \frac{t}{\sqrt{1+t^2}}\right) dt = \frac{1}{\sqrt{1+t^2}} dt$.$I = [t \ln(t+\sqrt{1+t^2})]_{1/2}^{2} - \int \limits_{1/2}^{2} \frac{t}{\sqrt{1+t^2}} dt$.$I = [2 \ln(2+\sqrt{5}) - \frac{1}{2} \ln(\frac{1}{2} + \sqrt{1 + 1/4})] - [\sqrt{1+t^2}]_{1/2}^{2}$.$I = 2 \ln(2+\sqrt{5}) - \frac{1}{2} \ln(\frac{1+\sqrt{5}}{2}) - (\sqrt{5} - \sqrt{5/4})$.$I = \ln((2+\sqrt{5})^2) - \ln((\frac{1+\sqrt{5}}{2})^{1/2}) - \frac{\sqrt{5}}{2}$.$I = \ln \left( \frac{(2+\sqrt{5})^2}{\sqrt{\frac{1+\sqrt{5}}{2}}} \right) - \frac{\sqrt{5}}{2} = \ln \left( \frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}} \right) - \frac{\sqrt{5}}{2}$.

$t \ (\text{in second})$	$0$	$2$	$4$	$6$	$8$	$10$
$v \ (\text{in m/s})$	$0$	$12$	$16$	$20$	$35$	$60$

The value of the integral $\int \limits_{-\log _{e} 2}^{\log _e 2} e^x \ln \left(e^x+\sqrt{1+e^{2 x}}\right) d x$ is equal to

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