If int frac{log left(t+sqrt{1+t^2}right)}{sqr | vedclass

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

Question

(D) Let $y = \log \left(t+\sqrt{1+t^2}\right)$.Then,differentiating with respect to $t$,we get $dy = \frac{1}{t+\sqrt{1+t^2}} \left(1 + \frac{2t}{2\sq

Vedclass · Accepted Answer

$\log (2+\sqrt{5})$

Vedclass · Answer

(D) Let $y = \log \left(t+\sqrt{1+t^2}\right)$.Then,differentiating with respect to $t$,we get $dy = \frac{1}{t+\sqrt{1+t^2}} \left(1 + \frac{2t}{2\sqrt{1+t^2}}\right) dt$.Simplifying the term inside the parenthesis: $1 + \frac{t}{\sqrt{1+t^2}} = \frac{\sqrt{1+t^2}+t}{\sqrt{1+t^2}}$.Thus,$dy = \frac{1}{t+\sqrt{1+t^2}} \cdot \frac{\sqrt{1+t^2}+t}{\sqrt{1+t^2}} dt = \frac{1}{\sqrt{1+t^2}} dt$.Substituting this into the integral: $\int y \, dy = \frac{y^2}{2} + c$.Comparing this with the given expression $\frac{1}{2}[g(t)]^2 + c$,we identify $g(t) = y = \log \left(t+\sqrt{1+t^2}\right)$.Therefore,$g(2) = \log \left(2+\sqrt{1+2^2}\right) = \log (2+\sqrt{5})$.

If $\int \frac{\log \left(t+\sqrt{1+t^2}\right)}{\sqrt{1+t^2}} dt=\frac{1}{2}[g(t)]^2+c$,(where $c$ is a constant of integration),then $g(2)$ is

Similar Questions

If a gas contains only three molecules that move with velocities of $100, 200, 500 \ ms^{-1}$. What is the $rms$ velocity of that gas in $ms^{-1}$?

If the curves $y^2=6x$ and $9x^2+by^2=16$ intersect each other at right angles,then the value of $b$ is

The transformed equation of $x^2+6xy+8y^2=10$ when the axes are rotated through an angle $\frac{\pi}{4}$ is:

In which of the following is the hydration energy higher than the lattice energy?

Which gas is evolved when $SiO_2$ reacts with sodium carbonate?

Vedclass Test Series

Exam Paper Generator

Online Exam Module