Question

What is the value of the definite integral \[\int_{ - 1}^1 {\log \left( {x + \sqrt {{x^2} + 1} } \right)dx} \]?
A. 0
B. \[\log 2\]
C. \[\log \dfrac{1}{2}\]
D. None of these

Answer 1

Hint: The integration has limit thus it is a definite integral. Then we will check whether the given function that is \[\log \left( {x + \sqrt {{x^2} + 1} } \right)\] is an odd function or an even function by putting x = -x. After that we will apply definite integral property to find the value of the integration.



Formula Used:Definite integral property:
\[\int\limits_{ - a}^a {f\left( x \right)dx} = \left\{ {\begin{array}{*{20}{c}}0&{{\rm{if f(x) is an odd function}}}\\{2\int_0^a {f\left( x \right)dx} }&{{\rm{if f(x) is an even function}}}\end{array}} \right.\]



Complete step by step solution:Given definite integral is \[\int_{ - 1}^1 {\log \left( {x + \sqrt {{x^2} + 1} } \right)dx} \].
Assume that, \[f\left( x \right) = \log \left( {x + \sqrt {{x^2} + 1} } \right)\]
To check whether\[f\left( x \right)\] is an odd function or an even function, we will put x = -x in \[f\left( x \right) = \log \left( {x + \sqrt {{x^2} + 1} } \right)\]:
\[f\left( { - x} \right) = \log \left( {\left( { - x} \right) + \sqrt {{{\left( { - x} \right)}^2} + 1} } \right)\]
\[ \Rightarrow f\left( { - x} \right) = \log \left( {\sqrt {{x^2} + 1} - x} \right)\]
Now multiplying and divide \[\left( {\sqrt {{x^2} + 1} + x} \right)\] with \[\left( {\sqrt {{x^2} + 1} - x} \right)\]
\[ \Rightarrow f\left( { - x} \right) = \log \left( {\left( {\sqrt {{x^2} + 1} - x} \right) \times \dfrac{{\left( {\sqrt {{x^2} + 1} + x} \right)}}{{\left( {\sqrt {{x^2} + 1} + x} \right)}}} \right)\]
\[ \Rightarrow f\left( { - x} \right) = \log \left( {\dfrac{{\left( {\sqrt {{x^2} + 1} - x} \right) \times \left( {\sqrt {{x^2} + 1} + x} \right)}}{{\left( {\sqrt {{x^2} + 1} + x} \right)}}} \right)\]
Applying algebraic identity \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\]
  \[ \Rightarrow f\left( { - x} \right) = \log \left( {\dfrac{{{{\left( {\sqrt {{x^2} + 1} } \right)}^2} - {x^2}}}{{\left( {\sqrt {{x^2} + 1} + x} \right)}}} \right)\]
\[ \Rightarrow f\left( { - x} \right) = \log \left( {\dfrac{{{x^2} + 1 - {x^2}}}{{\left( {\sqrt {{x^2} + 1} + x} \right)}}} \right)\]
\[ \Rightarrow f\left( { - x} \right) = \log \left( {\dfrac{1}{{\left( {\sqrt {{x^2} + 1} + x} \right)}}} \right)\]
Now applying the quotient formula of logarithm function \[\log \dfrac{a}{b} = \log a - \log b\]:
\[ \Rightarrow f\left( { - x} \right) = \log \left( 1 \right) - \log \left( {\sqrt {{x^2} + 1} + x} \right)\]
Now putting \[\log 1 = 0\]
\[ \Rightarrow f\left( { - x} \right) = - \log \left( {\sqrt {{x^2} + 1} + x} \right)\]
\[ \Rightarrow f\left( { - x} \right) = - f\left( x \right)\]
Thus \[f\left( x \right)\] is an odd function.
Now applying the property of the definite integral \[\int\limits_{ - a}^a {f\left( x \right)dx} = \left\{ {\begin{array}{*{20}{c}}0&{{\rm{if f(x) is an odd function}}}\\{2\int_0^a {f\left( x \right)dx} }&{{\rm{if f(x) is an even function}}}\end{array}} \right.\] in the given integration:
\[\int_{ - 1}^1 {\log \left( {x + \sqrt {{x^2} + 1} } \right)dx} = 0\]



Option ‘A’ is correct



Note: Students often follow a wrong way to solve the given definite integration. They apply by parts formula to solve it. It is a bit lengthy process. We can solve the given definite integration by using property of definite integration.