Question

Find the value of the given definite integral: $ \int\limits_{0}^{1}{\dfrac{\ln \left( 1+x \right)}{1+{{x}^{2}}}dx} $ ?
(a) $ \dfrac{\pi }{4}\ln 2 $
(b) $ \dfrac{\pi }{2}\ln 2 $
(c) $ \dfrac{\pi }{8}\ln 2 $
(d) $ \pi \ln 2 $

Answer 1

Hint: We start solving the problem by equating the given definite integer to a variable. We then assume $ x=\tan \theta $ and find the $ dx $ in terms of $ d\theta $ , upper limits and lower limits for the updated integral. We then make the necessary calculations and make use of the facts that $ \int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx} $ and $ \tan \left( \dfrac{\pi }{4}-\theta \right)=\dfrac{1-\tan \theta }{1+\tan \theta } $ to get the equivalent form of definite integral. We then add both the forms of integrals and make use of the facts $ \int{adx}=ax+C $ and $ \int\limits_{a}^{b}{{{f}^{'}}\left( x \right)dx}=\left[ f\left( x \right) \right]_{a}^{b}=f\left( b \right)-f\left( a \right) $ to get the required answer.

Complete step by step answer:
According to the problem, we need to find the value of the definite integral $ \int\limits_{0}^{1}{\dfrac{\ln \left( 1+x \right)}{1+{{x}^{2}}}dx} $ .
Let us assume $ I=\int\limits_{0}^{1}{\dfrac{\ln \left( 1+x \right)}{1+{{x}^{2}}}dx} $ ---(1).
Now, let us take $ x=\tan \theta $ ---(2) and apply the differential on both sides.
 $ \Rightarrow d\left( x \right)=d\left( \tan \theta \right) $ .
 $ \Rightarrow dx={{\sec }^{2}}\theta d\theta $ ---(3).
Now, we have a lower limit $ x=0 $ , then $ \theta ={{\tan }^{-1}}\left( 0 \right)\Leftrightarrow \theta =0 $ ---(4).
Now, we have lower limit $ x=1 $ , then $ \theta ={{\tan }^{-1}}\left( 1 \right)\Leftrightarrow \theta =\dfrac{\pi }{4} $ ---(5).
Let us substitute equations (2), (3), (4), (5) in equation (1).
 $ \Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{\ln \left( 1+\tan \theta \right)}{1+{{\tan }^{2}}\theta }\times {{\sec }^{2}}\theta d\theta } $ .
We know that $ 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $ .
 $ \Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{\ln \left( 1+\tan \theta \right)}{{{\sec }^{2}}\theta }\times {{\sec }^{2}}\theta d\theta } $ .
 $ \Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{4}}{\ln \left( 1+\tan \theta \right)d\theta } $ ---(6).
We know that $ \int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx} $ .
 $ \Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{4}}{\ln \left( 1+\tan \left( \dfrac{\pi }{4}-\theta \right) \right)d\theta } $ .
We know that $ \tan \left( \dfrac{\pi }{4}-\theta \right)=\dfrac{1-\tan \theta }{1+\tan \theta } $ .
 $ \Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{4}}{\ln \left( 1+\left( \dfrac{1-\tan \theta }{1+\tan \theta } \right) \right)d\theta } $ .
 $ \Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{4}}{\ln \left( \dfrac{1+\tan \theta +1-\tan \theta }{1+\tan \theta } \right)d\theta } $ .
 $ \Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{4}}{\ln \left( \dfrac{2}{1+\tan \theta } \right)d\theta } $ .
We know that $ \ln \left( \dfrac{a}{b} \right)=\ln a-\ln b $ .
 $ \Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( \ln \left( 2 \right)-\ln \left( 1+\tan \theta \right) \right)d\theta } $ .
We know that $ \int{\left( a-b \right)dx}=\int{adx}-\int{bdx} $ .
 $ \Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{4}}{\ln 2d\theta }-\int\limits_{0}^{\dfrac{\pi }{4}}{\ln \left( 1+\tan \theta \right)d\theta } $ ---(7).
Let us add equations (6) and (7).
 $ \Rightarrow I+I=\int\limits_{0}^{\dfrac{\pi }{4}}{\ln 2d\theta }-\int\limits_{0}^{\dfrac{\pi }{4}}{\ln \left( 1+\tan \theta \right)d\theta }+\int\limits_{0}^{\dfrac{\pi }{4}}{\ln \left( 1+\tan \theta \right)d\theta } $ .
 $ \Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{4}}{\ln 2d\theta } $ .
We know that $ \int{adx}=ax+C $ and $ \int\limits_{a}^{b}{{{f}^{'}}\left( x \right)dx}=\left[ f\left( x \right) \right]_{a}^{b}=f\left( b \right)-f\left( a \right) $ .
 $ \Rightarrow 2I=\left[ x\ln 2 \right]_{0}^{\dfrac{\pi }{4}} $ .
 $ \Rightarrow 2I=\left( \dfrac{\pi }{4}\ln 2 \right)-\left( 0\ln 2 \right) $ .
 $ \Rightarrow 2I=\dfrac{\pi }{4}\ln 2 $ .
 $ \Rightarrow I=\dfrac{\pi }{8}\ln 2 $ .
 $ \Rightarrow \int\limits_{0}^{1}{\dfrac{\ln \left( 1+x \right)}{1+{{x}^{2}}}dx}=\dfrac{\pi }{8}\ln 2 $ .
 $ \therefore, $ The correct option for the given problem is (c).

Note:
We can see that the given problem contains a huge amount of calculation, so we need to perform each step carefully in order to avoid confusion and calculation mistakes. We should not forget to change the limits of the integral after taking $ x=\tan \theta $ as this is the most common mistake done by mistakes. We should not forget to divide the answer after finding the value of 2I. Similarly, we can expect problems to find the value of the definite integral $ \int\limits_{0}^{\infty }{\dfrac{\ln \left( 1+x \right)}{1+{{x}^{2}}}dx} $ .