Question

The integral \[\int_0^{\dfrac{1}{2}} {\dfrac{{\ln \left( {1 + 2x} \right)}}{{1 + 4{x^2}}}dx} \] equals:
A) \[\dfrac{\pi }{4}\ln 2\]
B) \[\dfrac{\pi }{8}\ln 2\]
C) \[\dfrac{\pi }{{12}}\ln 2\]
D) \[\dfrac{\pi }{{32}}\ln 2\]

Answer 1

Hint: Here first we will assume \[2x = y\]and then substitute the values of limits so obtained as well the value of the expression inside the integral. Then we will use the following identity:-
 \[1 + {\tan ^2}\theta = {\sec ^2}\theta \]
And again substitute the values of limits and the expression inside the integral and finally then we will use the property of definite integrals i.e. \[\int_a^b {f\left( x \right)dx = \int_a^b {f\left( {a + b - x} \right)} } dx\] and then solve the integral to get the desired answer.

Complete step-by-step solution:
The given expression can be written as:-
 \[ = \int_0^{\dfrac{1}{2}} {\dfrac{{\ln \left( {1 + 2x} \right)}}{{1 + {{\left( {2x} \right)}^2}}}dx}
\]………………………………………………….(1)
Now let us assume \[2x = y\]
Differentiating both the sides we get:-
 \[2\left( {\dfrac{d}{{dx}}\left( x \right)} \right) = \dfrac{d}{{dx}}\left( y \right)\]
Applying the following formula:-
 \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
We get:-
 \[
2 \times {x^0} = \dfrac{{dy}}{{dx}} \\
\Rightarrow 2 = \dfrac{{dy}}{{dx}}
\]
Evaluating the value of \[dx\] we get:-
 \[
2dx = dy \\
\Rightarrow dx = \dfrac{{dy}}{2}
\]
Then putting \[x = 0\] as lower limit we get
 \[
y = 2\left( 0 \right) \\
y = 0
\]
Now putting \[x = \dfrac{1}{2}\] as upper limit we get:-
 \[
y = 2\left( {\dfrac{1}{2}} \right) \\
\Rightarrow y = 1
\]
Hence the new limits are:-
Lower limit- \[y = 0\]
Upper limit - \[y = 1\]
Hence substituting the new limits and \[2x = y\] and \[dx\] in equation 1 we get:-
 \[
= \int_0^1 {\dfrac{{\ln \left( {1 + y} \right)}}{{1 + {y^2}}}\left( {\dfrac{{dy}}{2}} \right)} \\
= \dfrac{1}{2}\int_0^1 {\dfrac{{\ln \left( {1 + y} \right)}}{{1 + {y^2}}}dy} ........................................\left( 2\right)
\]
Now substitute \[y = \tan \theta \]
Differentiating both the sides we get:-
 \[\dfrac{d}{{dy}}\left( y \right) = \dfrac{d}{{d\theta }}\left( {\tan \theta } \right)\]
Applying the following formulas:-
 \[
\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} \\
\dfrac{d}{{dx}}\tan x = {\sec ^2}x
\]
We get:-
 \[
{y^0} = {\sec ^2}\theta .\dfrac{{d\theta }}{{dy}} \\
\Rightarrow 1 = {\sec ^2}\theta .\dfrac{{d\theta }}{{dy}}
\]
Evaluating the value of \[dy\] we get:-
 \[dy = {\sec ^2}\theta d\theta \]
Then putting \[y = 0\] as lower limit we get
 \[\tan \theta = 0\]
Now we know that \[\tan 0 = 0\]
Hence, we get \[\theta = 0\]
Now putting \[y = 1\] as upper limit we get:-
 \[\tan \theta = 1\]
Now we know that \[\tan \left( {\dfrac{\pi }{4}} \right) = 1\]
Hence, we get \[\theta = \dfrac{\pi }{4}\]
Hence the new limits are:-
Lower limit- \[\theta = 0\]
Upper limit - \[\theta = \dfrac{\pi }{4}\]
Hence substituting the new limits and \[y = \tan \theta \] and \[dy\] in equation 2 we get:-

 \[ = \dfrac{1}{2}\int_0^{\dfrac{\pi }{4}} {\dfrac{{\ln \left( {1 + \tan \theta } \right)}}{{1 + {{\left( {\tan\theta } \right)}^2}}}} {\sec ^2}\theta d\theta \]
Now we know that:
 \[1 + {\left( {\tan \theta } \right)^2} = {\sec ^2}\theta \]
Hence substituting the value in above expression we get:-
 \[ = \dfrac{1}{2}\int_0^{\dfrac{\pi }{4}} {\dfrac{{\ln \left( {1 + \tan \theta } \right)}}{{{{\sec }^2}\theta }}}{\sec ^2}\theta d\theta \]
Now cancelling the required terms we get:-
 \[ = \dfrac{1}{2}\int_0^{\dfrac{\pi }{4}} {\ln \left( {1 + \tan \theta } \right)} d\theta \]
Let,
 \[I = \int_0^{\dfrac{\pi }{4}} {\ln \left( {1 + \tan \theta } \right)} d\theta \]………………………………….(3)
Now we will apply the following property of definite integrals:
 \[\int_a^b {f\left( x \right)dx = \int_a^b {f\left( {a + b - x} \right)} } dx\]
Hence on applying this property we get:-
$\Rightarrow$ \[I = \dfrac{1}{2}\int_0^{\dfrac{\pi }{4}} {\ln \left( {1 + \tan \left( {\dfrac{\pi }{4} + 0 - \theta } \right)}\right)} d\theta \]
Simplifying it further we get:-
$\Rightarrow$ \[I = \dfrac{1}{2}\int_0^{\dfrac{\pi }{4}} {\ln \left( {1 + \tan \left( {\dfrac{\pi }{4} - \theta } \right)} \right)}d\theta \]
Now we will use the following identity:-
 \[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\]
Applying this identity in above expression we get:-
$\Rightarrow$ \[I = \dfrac{1}{2}\int_0^{\dfrac{\pi }{4}} {\ln \left( {1 + \dfrac{{\tan \dfrac{\pi }{4} - \tan \theta }}{{1 + \tan\dfrac{\pi }{4}.\tan \theta }}} \right)} d\theta \]
Now we know that:-
 \[\tan \left( {\dfrac{\pi }{4}} \right) = 1\]
Hence putting the value we get:-
 \[
I = \dfrac{1}{2}\int_0^{\dfrac{\pi }{4}} {\ln \left( {1 + \dfrac{{1 - \tan \theta }}{{1 + \left( 1 \right).\tan\theta }}} \right)} d\theta \\
\Rightarrow I = \dfrac{1}{2}\int_0^{\dfrac{\pi }{4}} {\ln \left( {1 + \dfrac{{1 - \tan \theta }}{{1 + \tan\theta }}} \right)} d\theta \\
\]
Simplifying it further and taking the LCM we get:-
$\Rightarrow$ \[I = \dfrac{1}{2}\int_0^{\dfrac{\pi }{4}} {\ln \left( {\dfrac{{1 + \tan \theta + 1 - \tan \theta }}{{1 + \tan\theta }}} \right)} d\theta \]
Cancelling the required terms we get:-
 \[
\Rightarrow I = \dfrac{1}{2}\int_0^{\dfrac{\pi }{4}} {\ln \left( {\dfrac{{1 + 1}}{{1 + \tan \theta }}} \right)} d\theta
\\
\Rightarrow I = \dfrac{1}{2}\int_0^{\dfrac{\pi }{4}} {\ln \left( {\dfrac{2}{{1 + \tan \theta }}} \right)}d\theta
\]
Now using the following property of log function:-
 \[\log \left( {\dfrac{a}{b}} \right) = \log \left( a \right) - \log \left( b \right)\]
We get:-
$\Rightarrow$ \[I = \dfrac{1}{2}\int_0^{\dfrac{\pi }{4}} {\ln \left( 2 \right) - \ln \left( {1 + \tan \theta } \right)} d\theta \]
Now distributing the integral for both the terms we get:-
$\Rightarrow$ \[I = \dfrac{1}{2}\int_0^{\dfrac{\pi }{4}} {\ln \left( 2 \right)d\theta - \dfrac{1}{2}\int_0^{\dfrac{\pi }{4}}{\ln \left( {1 + \tan \theta } \right)} } d\theta \]…………………………………….(4)
Now from equation 3,
$\Rightarrow$ \[I = \int_0^{\dfrac{\pi }{4}} {\ln \left( {1 + \tan \theta } \right)} d\theta \]
Hence substituting this value in equation 4 we get:-
$\Rightarrow$ \[I = \dfrac{1}{2}\int_0^{\dfrac{\pi }{4}} {\ln \left( 2 \right)d\theta - \dfrac{I}{2}} \]
Simplifying it further we get:-
 \[
I + \dfrac{I}{2} = \dfrac{1}{2}\int_0^{\dfrac{\pi }{4}} {\ln \left( 2 \right)d\theta } \\
\Rightarrow \dfrac{{3I}}{2} = \dfrac{{\ln \left( 2 \right)}}{2}\int_0^{\dfrac{\pi }{4}} {d\theta }
\]
Now using the following formula for integration:-
 \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + C\]
We get:-
$\Rightarrow$ \[3I = \ln \left( 2 \right)\left[ \theta \right]_0^{\dfrac{\pi }{4}}\]
Now putting the upper limit and lower limit we get:-
$\Rightarrow$ \[3I = \ln \left( 2 \right)\left[ {\dfrac{\pi }{4} - 0} \right]\]
Simplifying it further we get:-
 \[
3I = \dfrac{\pi }{4}\ln \left( 2 \right) \\
\Rightarrow I = \dfrac{\pi }{{12}}\ln \left( 2 \right) \\
\]

Hence option C is the correct answer.

Note: Students might make mistakes in using the various formulas and identities used so all the formulas should be correct and they might forget to change the limits according to the substitution done.
Also, students should note that the general formula for integration is:
 \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + C\]