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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]4 Marks
Question
Evaluate: $\int_0^1 \frac{\log (x+1)}{x^2+1} d x$
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Answer

Let $I =\int_0^1 \frac{\log (x+1)}{x^2+1} d x$
Put $x=\tan \theta$
$\therefore dx =\sec ^2 \theta d \theta$
When $x =0, \theta=0$ and when $x =1, \theta=\frac{\pi}{4}$
$ \therefore I =\int_0^{\frac{\pi}{4}} \frac{\log (\tan \theta+1)}{\tan ^2 \theta+1} \times \sec ^2 \theta d \theta$
$=\int_0^{\frac{\pi}{4}} \frac{\log (1+\tan \theta)}{\sec ^2 \theta} \times \sec ^2 \theta d \theta$
$\therefore I =\int_0^{\frac{\pi}{4}} \log (1+\tan \theta) d \theta \ldots . . \text { (i) }$
$=\int_0^{\frac{\pi}{4}} \log \left[1+\tan \left(\frac{\pi}{4}-\theta\right)\right] d \theta \quad \ldots . .\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right] $
$=\int_0^{\frac{\pi}{4}} \log \left[1+\frac{\tan \frac{\pi}{4}-\tan \theta}{1+\tan \frac{\pi}{4} \tan \theta}\right] d \theta$
$=\int_0^{\frac{\pi}{4}} \log \left[1+\frac{1-\tan \theta}{1+\tan \theta}\right] d \theta$
$=\int_0^{\frac{\pi}{4}} \log \left[\frac{1+\tan \theta+1-\tan \theta}{1+\tan \theta}\right] d \theta$
$=\int_0^{\frac{\pi}{4}} \log \left[\frac{2}{1+\tan \theta}\right] d \theta$
$=\int_0^{\frac{\pi}{4}}[\log 2-\log (1+\tan \theta)] d \theta$
$=\log 2 \int_0^{\frac{\pi}{4}} d \theta-\int_0^{\frac{\pi}{4}} \log (1+\tan \theta) d \theta$
$\therefore I =\log 2[\theta]_0^{\frac{\pi}{4}}- I \ldots .[\text { From (i) }]$
$\therefore 2 I =\log 2\left(\frac{\pi}{4}-0\right)$
$\therefore I =\frac{\pi}{8} \log 2$

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