Question

If $\int {\dfrac{{\log (t + \sqrt {1 + {t^2}} )}}{{\sqrt {1 + {t^2}} }}dt = \dfrac{1}{2}{{(g(t))}^2} + C}$ where $C$ is a constant, then $g(2)$ is equal to:
$A)\,2\,\log (2 + \sqrt 5 ) \\\ B)\,\log (2 + \sqrt 5 ) \\\ C)\,\dfrac{1}{{\sqrt 5 }}\,\log (2 + \sqrt 5 ) \\\ D)\,\dfrac{1}{2}\,\,\log (2 + \sqrt 5 ) \\\$

Answer 1

The given function contains a logarithmic part in it. Split the logarithmic part and assign any variable to it. By using a logarithmic formula find the value of it. And substitute the obtained value in the given function and use integral and differential formula to get an accurate answer of given $g(2)$.

Formula used: Integral formula of $\int {u\,du = \dfrac{{{u^2}}}{2} + C}$ as with respect to $'u'$.

Complete step-by-step answer:
Given that $I = \int {\dfrac{{\log (t + \sqrt {1 + {t^2}} )}}{{\sqrt {1 + {t^2}} }}\,dt = \dfrac{1}{2}{{(g(t))}^2}} + C$. The Integral Equation is given with value.
Split the above equation in two parts like logarithmic part and differentiable part which as follows:
$I = \int {\log (t + \sqrt {1 + {t^2}} )\, \cdot \,\dfrac{1}{{\sqrt {1 + {t^2}} }}dt}$
Now, let us assign $u = \log (t + \sqrt {1 + {t^2}} )$
Differentiate the $u$ with respect to $'t'$, the equation becomes
$\Rightarrow du = \dfrac{1}{{t + \sqrt {1 + {t^2}} }}(1 + \dfrac{{2t}}{{2\sqrt {1 + {t^2}} }})dt$
Cancel the value $2$ in both numerator and denominator the equation becomes,
$\Rightarrow du = \dfrac{1}{{t + \sqrt {1 + {t^2}} }}(1 + \dfrac{t}{{\sqrt {1 + {t^2}} }})dt$
Now take LCM (Least Common Multiplier) of $\sqrt {1 + {t^2}}$ in $(1 + \dfrac{t}{{\sqrt {1 + {t^2}} }})$, the equation should be as follows $\Rightarrow du = \dfrac{1}{{t + \sqrt {1 + {t^2}} }}(\dfrac{{\sqrt {1 + {t^2}} }}{{\sqrt {1 + {t^2}} }} + \dfrac{t}{{\sqrt {1 + {t^2}} }})dt$
While the denominator is the same, take it as a common denominator to all nominators.
By using this condition take $\sqrt {1 + {t^2}}$ as the common part. The equation should be as:
$\Rightarrow du = \dfrac{1}{{t + \sqrt {1 + {t^2}} }}(\dfrac{{\sqrt {1 + {t^2}} + t}}{{\sqrt {1 + {t^2}} }})\,dt$
Cancel the similar parts present in the numerator and denominator, thus the equation changes as
$\therefore du = \dfrac{1}{{\sqrt {1 + {t^2}} }}dt$
Finally, we determine the value of $u$ by using differential equations:
Thus, $u = \log (t + \sqrt {1 + {t^2}} )$ and $\dfrac{1}{{\sqrt {1 + {t^2}} }}dt = du$
From, the question $I = \int {\dfrac{{\log (t + \sqrt {1 + {t^2}} )}}{{\sqrt {1 + {t^2}} }}} dt$
The above equation should be split as follows:

By using the values of $u$ and $du$ in above equation, we get
$I = \int {u\,du}$
Now, use the Integral formula in $I$
$\Rightarrow \int {u\,du = \dfrac{{{u^2}}}{2}} + C$
Substitute the value of $u$in the above equation.
$\Rightarrow \int {u\,du = \dfrac{1}{2}\,\log (t + \sqrt {1 + {t^2}} ) + C}$
Compare the above equation with the given solution of $\dfrac{1}{2}{(g(t))^2}$ to find the value of $g(t)$.
$\Rightarrow \dfrac{1}{2}{(g(t))^2} = \dfrac{1}{2}{(\log (t + \sqrt {1 + {t^2}} ))^2}$
Cancel the common parts of $\dfrac{1}{2}$ and $C$ from both the left and right side of the equation. Then the equation becomes as follows:
$\Rightarrow {(g(t))^2} = {(\log (t + \sqrt {1 + {t^2}} ))^2}$
Taking the square root on each side, the equation is
$\Rightarrow g(t) = \log (t + \sqrt {1 + {t^2}} )$
Now, Substitute the value of $t = 2$ in the above equation:
$\Rightarrow g(2) = \log (2 + \sqrt {1 + {2^2}} )$
Squaring the value $2$ to simplify the equation:
$\Rightarrow g(2) = \log (2 + \sqrt 5 )$
$\therefore$ The value for $g(2)$ is $\log (2 + \sqrt 5 )$.

So, the correct answer is “Option B”.

Note: If the function contains the logarithmic part in it, split the equation by two parts and if possible try to use a substitution method for solving the integration to make the problem easier. Simplify the given equation as possible before integrating the function, to reduce the math error in the problem.