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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $x = -\ln 2$,$

Vedclass · Accepted Answer

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

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 $du = \frac{1}{t+\sqrt{1+t^2}} \left(1 + \frac{t}{\sqrt{1+t^2}}\right) dt = \frac{1}{\sqrt{1+t^2}} dt$.$I = [t \ln(t+\sqrt{1+t^2})]_{1/2}^{2} - \int \limits_{1/2}^{2} \frac{t}{\sqrt{1+t^2}} dt$.$I = [2 \ln(2+\sqrt{5}) - \frac{1}{2} \ln(\frac{1}{2} + \sqrt{1 + 1/4})] - [\sqrt{1+t^2}]_{1/2}^{2}$.$I = 2 \ln(2+\sqrt{5}) - \frac{1}{2} \ln(\frac{1+\sqrt{5}}{2}) - (\sqrt{5} - \sqrt{5/4})$.$I = \ln((2+\sqrt{5})^2) - \ln((\frac{1+\sqrt{5}}{2})^{1/2}) - \frac{\sqrt{5}}{2}$.$I = \ln \left( \frac{(2+\sqrt{5})^2}{\sqrt{\frac{1+\sqrt{5}}{2}}} \right) - \frac{\sqrt{5}}{2} = \ln \left( \frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}} \right) - \frac{\sqrt{5}}{2}$.

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

Similar Questions

If $\int_{\log _{e} 2}^{x} (e^{t}-1)^{-1} dt = \log _{e} \frac{3}{2}$, then the value of $x$ is

If the integral $525 \int_0^{\frac{\pi}{2}} \sin 2 x \cos^{\frac{11}{2}} x \left(1+\cos^{\frac{5}{2}} x\right)^{\frac{1}{2}} d x$ is equal to $(n \sqrt{2}-64)$,then $n$ is equal to

$\int\limits_0^{\frac{\pi }{2}} \frac{dx}{1 + a^2 \sin^2 x}$ has the value :

Let $\phi (x) = \int_{0}^{1} e^{x} e^{t} \phi (t) dt + x$. If $\phi (\ln (e^{2} - 3))$ is equal to $A$,then find the value of $A$.

Let $[t]$ denote the largest integer less than or equal to $t$. If $\int_0^3 \left( [x^2] + [\frac{x^2}{2}] \right) dx = a + b\sqrt{2} - \sqrt{3} - \sqrt{5} + c\sqrt{6} - \sqrt{7}$,where $a, b, c \in \mathbb{Z}$,then $a + b + c$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module