Question
Evaluate the following integrals:$\int\limits^{1}_0\frac{\log(1+\text{x})}{1+\text{x}^2}\text{ dx}$
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Answer
We have,$\text{I}=\int\limits^{1}_0\frac{\log(1+\text{x})}{1+\text{x}^2}\text{ dx}$
Putting $\text{x}=\tan\theta$$\text{dx}=\sec^2\theta\text{ d}\theta$
When $\text{x}\rightarrow0;\theta\rightarrow0$ And $\text{x}\rightarrow1;\theta\rightarrow\frac{\pi}{4}$ Now, integral becomes,$\text{I}=\int\limits^{\frac{\pi}{4}}_0\frac{\log(1+\tan\theta)}{\sec^2\theta}\sec^2\theta\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\big[\log(\tan\theta)\big]\text{d}\theta\ ...(\text{i})$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\bigg[\log\Big\{1+\tan\Big(\frac{\pi}{4}-\theta\Big)\Big\}\bigg]\text{d}\theta$ $\Bigg[\because\ \int\limits^{\text{a}}_{0}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_{\text{0}}\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\Bigg[\log\bigg\{1+\frac{\tan\frac{\pi}{4}-\tan\theta}{1+\tan\frac{\pi}{4}\tan\theta}\bigg\}\Bigg]\text{d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\bigg[\log\Big\{1+\frac{1-\tan\theta}{1+\tan\theta}\Big\}\bigg]\text{d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\bigg[\log\Big\{\frac{2}{1+\tan\theta}\Big\}\bigg]\text{d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\big[\log2-\log(1+\tan\theta)\big]\text{d}\theta\ ...(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\frac{\pi}{4}}_0\big(\log2\big)\text{d}\theta$
$\Rightarrow2\text{I}=\big(\log2\big)\Big[\theta\Big]^{\frac{\pi}{4}}_0$
$\Rightarrow2\text{I}=\frac{\pi}{4}\log2$
$\Rightarrow\text{I}=\frac{\pi}{8}\log2$
$\therefore\ \int\limits^{1}_0\frac{\log(1+\text{x})}{1+\text{x}^2}\text{ dx}=\frac{\pi}{8}\log2$
Putting $\text{x}=\tan\theta$$\text{dx}=\sec^2\theta\text{ d}\theta$
When $\text{x}\rightarrow0;\theta\rightarrow0$ And $\text{x}\rightarrow1;\theta\rightarrow\frac{\pi}{4}$ Now, integral becomes,$\text{I}=\int\limits^{\frac{\pi}{4}}_0\frac{\log(1+\tan\theta)}{\sec^2\theta}\sec^2\theta\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\big[\log(\tan\theta)\big]\text{d}\theta\ ...(\text{i})$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\bigg[\log\Big\{1+\tan\Big(\frac{\pi}{4}-\theta\Big)\Big\}\bigg]\text{d}\theta$ $\Bigg[\because\ \int\limits^{\text{a}}_{0}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_{\text{0}}\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\Bigg[\log\bigg\{1+\frac{\tan\frac{\pi}{4}-\tan\theta}{1+\tan\frac{\pi}{4}\tan\theta}\bigg\}\Bigg]\text{d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\bigg[\log\Big\{1+\frac{1-\tan\theta}{1+\tan\theta}\Big\}\bigg]\text{d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\bigg[\log\Big\{\frac{2}{1+\tan\theta}\Big\}\bigg]\text{d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\big[\log2-\log(1+\tan\theta)\big]\text{d}\theta\ ...(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\frac{\pi}{4}}_0\big(\log2\big)\text{d}\theta$
$\Rightarrow2\text{I}=\big(\log2\big)\Big[\theta\Big]^{\frac{\pi}{4}}_0$
$\Rightarrow2\text{I}=\frac{\pi}{4}\log2$
$\Rightarrow\text{I}=\frac{\pi}{8}\log2$
$\therefore\ \int\limits^{1}_0\frac{\log(1+\text{x})}{1+\text{x}^2}\text{ dx}=\frac{\pi}{8}\log2$
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