Question

Find the value of \[\int_{ - 1}^1 {\log \dfrac{{2 - x}}{{2 + x}}} dx\].
A. \[2\]
B. \[1\]
C. \[ - 1\]
D. \[0\]

Answer 1

Hint: In this question, we have \[\int_{ - 1}^1 {\log \dfrac{{2 - x}}{{2 + x}}} dx\]. We will first determine the function \[f(x)\] is even or odd. And for this we will substitute \[ - x\]into the function in place of\[x\].
If we determined that the function is even, then we will use \[\int_{ - a}^a {f\left( x \right)dx = 2} \int_0^a {f\left( x \right)dx} \] identity.
If we determined that the function is odd, then we will use \[\int_{ - a}^a {f\left( x \right)dx = 0} \] identity.



Formula Used:1) In order to determine the function\[f(x)\] is even or odd, we will first substitute \[ - x\]into the function in place of\[x\]. The function is even if \[f(x)\]is obtained in its original form. If the original function is obtained in the negative form, the function is odd.
In other words, we can say that:
If \[f( - x) = f(x)\], the function is even
If \[f( - x) = - f(x)\], the function is odd
2) Integrals for Even/Odd functions:
\[\int_{ - a}^a {f\left( x \right)dx = 2} \int_0^a {f\left( x \right)dx} \], if\[f(x)\] is even function
\[\int_{ - a}^a {f\left( x \right)dx = 0} \], if \[f(x)\] is odd function



Complete step by step solution:In this problem, the integral is given as \[\int_{ - 1}^1 {\log \dfrac{{2 - x}}{{2 + x}}} dx\].
Let us consider \[f\left( x \right) = \log \dfrac{{2 - x}}{{2 + x}} = \log \left( {2 - x} \right) - \log \left( {2 + x} \right)\]
In order to determine the integral function is even or odd, we will substitute \[ - x\] into the function in place of \[x\].
\[f\left( { - x} \right) = \log \dfrac{{2 - \left( { - x} \right)}}{{2 + \left( { - x} \right)}} = \log \dfrac{{2 + x}}{{2 - x}}\]
\[f\left( { - x} \right) = \log \left( {2 + x} \right) - \log \left( {2 - x} \right)\]
\[f\left( { - x} \right) = - \left[ {\log \left( {2 - x} \right) - \log \left( {2 + x} \right)} \right]\]
We can see that \[f\left( { - x} \right) = - f\left( x \right)\], this implies that function \[f\left( x \right)\]is an odd function.
Now we will use the formula \[\int_{ - a}^a {f\left( x \right)dx = 0} \] because function\[f(x)\] is odd function.
So, we get
\[\int_{ - 1}^1 {\log \dfrac{{2 - x}}{{2 + x}}} dx = 0\]



Option ‘D’ is correct



Note: We must avoid errors and confusion while dealing with sign changes that occurred during integration. We need to adopt the identities that make the problems easier. Students are advised to write the correct identity and then solve the problem.