
If \[\int {\dfrac{{\log (t + \sqrt {1 + {t^2}} )}}{{\sqrt {1 + {t^2}} }}} dt = \dfrac{1}{2}{(g(t))^2} + c\] where c is constant, then \[g(2)\] is equal to,
A) \[\dfrac{1}{{\sqrt 5 }}\log (2 + \sqrt 5 ){\text{ }}\]
B) \[2\log (2 + \sqrt 5 ){\text{ }}\]
C) \[\log (2 + \sqrt 5 ){\text{ }}\]
D) \[\dfrac{1}{2}\log (2 + \sqrt 5 ){\text{ }}\]
Answer
510.6k+ views
Hint: Calculate the given integration using substitution method of the given function, we substitute \[x = \log (t + \sqrt {1 + {t^2}} )\]. And then on comparing both the sides find the value of \[g\left( t \right)\]. As the value of \[g(2)\] can be calculated from there by replacing the value of t with two.
Complete step by step solution: As the given integration is of \[\int {\dfrac{{\log (t + \sqrt {1 + {t^2}} )}}{{\sqrt {1 + {t^2}} }}} .dt\]
So, by using substitution method,
\[x = \log (t + \sqrt {1 + {t^2}} )\]
Now, differentiate both sides and then substitute them in the above given equation as,
\[
dx = \dfrac{1}{{t + \sqrt {1 + {t^2}} }}(1 + \dfrac{{2t}}{{2\sqrt {1 + {t^2}} }}).dt \\
dx = \dfrac{1}{{t + \sqrt {1 + {t^2}} }}(1 + \dfrac{t}{{\sqrt {1 + {t^2}} }}).dt \\
\]
Now take L.C.M from the above equation and simplifying it as,
\[
dx = \dfrac{1}{{t + \sqrt {1 + {t^2}} }}(\dfrac{{\sqrt {1 + {t^2}} }}{{\sqrt {1 + {t^2}} }} + \dfrac{t}{{\sqrt {1 + {t^2}} }}).dt \\
dx = \dfrac{1}{{t + \sqrt {1 + {t^2}} }}(\dfrac{{\sqrt {1 + {t^2}} + t}}{{\sqrt {1 + {t^2}} }}).dt \\
\]
As cancelling the common terms, we get,
\[dx = (\dfrac{1}{{\sqrt {1 + {t^2}} }}).dt\]
Hence, replacing all the substitution in the above integration \[\int {\dfrac{{\log (t + \sqrt {1 + {t^2}} )}}{{\sqrt {1 + {t^2}} }}} .dt\], we get,
\[I = \int x .dx\]
Hence, integrate the above term using general formula as,
\[ \Rightarrow I = \dfrac{{{x^2}}}{2} + c\]
And on comparing it with R.H.S we can conclude that
\[ \Rightarrow I = \dfrac{{{x^2}}}{2} + c = \dfrac{1}{2}{(g(t))^2} + c\]
So, the value of \[g\left( t \right)\] is given as,
\[ \Rightarrow g(t) = x = \log (t + \sqrt {1 + {t^2}} )\]
Hence, now calculate the value of \[g(2)\] as replacing the value of t in the above equation,
\[
\Rightarrow g(2) = \log (2 + \sqrt {1 + {2^2}} ) \\
\Rightarrow g(2) = \log (2 + \sqrt 5 ) \\
\]
Hence, option (C) is our required correct answer.
Note: The substitution method is used when an integral contains some function and its derivative. In this case, we can set u equal to the function and rewrite the integral in terms of the new variable u. This makes the integral easier to solve.
Also apply the differentiation properly as \[\dfrac{{d(\log x)}}{{dx}} = \dfrac{1}{x}\] and properly apply chain rule while further differentiation our given terms as \[\dfrac{{d(\sqrt x )}}{{dx}} = \dfrac{1}{{2\sqrt x }}\].
Complete step by step solution: As the given integration is of \[\int {\dfrac{{\log (t + \sqrt {1 + {t^2}} )}}{{\sqrt {1 + {t^2}} }}} .dt\]
So, by using substitution method,
\[x = \log (t + \sqrt {1 + {t^2}} )\]
Now, differentiate both sides and then substitute them in the above given equation as,
\[
dx = \dfrac{1}{{t + \sqrt {1 + {t^2}} }}(1 + \dfrac{{2t}}{{2\sqrt {1 + {t^2}} }}).dt \\
dx = \dfrac{1}{{t + \sqrt {1 + {t^2}} }}(1 + \dfrac{t}{{\sqrt {1 + {t^2}} }}).dt \\
\]
Now take L.C.M from the above equation and simplifying it as,
\[
dx = \dfrac{1}{{t + \sqrt {1 + {t^2}} }}(\dfrac{{\sqrt {1 + {t^2}} }}{{\sqrt {1 + {t^2}} }} + \dfrac{t}{{\sqrt {1 + {t^2}} }}).dt \\
dx = \dfrac{1}{{t + \sqrt {1 + {t^2}} }}(\dfrac{{\sqrt {1 + {t^2}} + t}}{{\sqrt {1 + {t^2}} }}).dt \\
\]
As cancelling the common terms, we get,
\[dx = (\dfrac{1}{{\sqrt {1 + {t^2}} }}).dt\]
Hence, replacing all the substitution in the above integration \[\int {\dfrac{{\log (t + \sqrt {1 + {t^2}} )}}{{\sqrt {1 + {t^2}} }}} .dt\], we get,
\[I = \int x .dx\]
Hence, integrate the above term using general formula as,
\[ \Rightarrow I = \dfrac{{{x^2}}}{2} + c\]
And on comparing it with R.H.S we can conclude that
\[ \Rightarrow I = \dfrac{{{x^2}}}{2} + c = \dfrac{1}{2}{(g(t))^2} + c\]
So, the value of \[g\left( t \right)\] is given as,
\[ \Rightarrow g(t) = x = \log (t + \sqrt {1 + {t^2}} )\]
Hence, now calculate the value of \[g(2)\] as replacing the value of t in the above equation,
\[
\Rightarrow g(2) = \log (2 + \sqrt {1 + {2^2}} ) \\
\Rightarrow g(2) = \log (2 + \sqrt 5 ) \\
\]
Hence, option (C) is our required correct answer.
Note: The substitution method is used when an integral contains some function and its derivative. In this case, we can set u equal to the function and rewrite the integral in terms of the new variable u. This makes the integral easier to solve.
Also apply the differentiation properly as \[\dfrac{{d(\log x)}}{{dx}} = \dfrac{1}{x}\] and properly apply chain rule while further differentiation our given terms as \[\dfrac{{d(\sqrt x )}}{{dx}} = \dfrac{1}{{2\sqrt x }}\].
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