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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
The value of the integral $\int\limits^\infty_0\frac{\text{x}}{(1+\text{x})(1+\text{x}^2)}\text{dx}$ is:
  • A
    $\frac{\pi}{2}$
  • ✓
    $\frac{\pi}{4}$
  • C
    $\frac{\pi}{6}$
  • D
    $\frac{\pi}{3}$
✓

Answer

Correct option: B.
$\frac{\pi}{4}$
We have,
$\text{I}=\int\limits^\infty_0\frac{\text{x}}{(1+\text{x})(1+\text{x}^2)}\text{ dx}$
Putting $\text{x}=\tan\theta$
$\Rightarrow \text{dx}=\sec^2\theta\text{ d}\theta$
when $\text{x}\rightarrow0;\theta\rightarrow0$
and $\text{x}\rightarrow\infty;\theta\rightarrow\frac{\pi}{2}$
Now, integral becomes
$\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\tan\theta}{(1+\tan\theta)\sec^2\theta}\sec^2\theta\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\tan\theta}{1+\tan\theta}\text{d}\theta$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\frac{\sin\theta}{\cos\theta}}{1+\frac{\sin\theta}{\cos\theta}}\text{d}\theta$
$\Rightarrow \text{I}=\int\limits^\frac{\pi}{2}_0\frac{\sin\theta}{\sin\theta+\cos\theta}\text{d}\theta\ ...(\text{i})$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\sin\big(\frac{\pi}{2}-\theta\big)}{\sin\big(\frac{\pi}{2}-\theta\big)+\cos\big(\frac{\pi}{2}-\theta\big)}\text{ d}\theta$ $\bigg[\therefore\ \int\limits^\text{a}_0\text{f}(\text{x)}\text{dx}=\int\limits^\text{a}_0\text{f}(\text{a}-\text{x})\text{dx}\bigg]$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\cos\theta}{\cos\theta+\sin\theta}\text{d}\theta$
$\Rightarrow \text{I}=\int\limits^\frac{\pi}{2}_0\frac{\cos\theta}{\sin\theta+\cos\theta}\text{d}\theta\ ....(\text{ii})$
Adding $(i)$ and $(ii),$ we get
$2\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\sin\theta+\cos\theta}{\sin\theta+\cos\theta}\text{d}\theta$
$\Rightarrow 2\text{I}=\int\limits^\frac{\pi}{2}_0\text{d}\theta$
$\Rightarrow 2\text{I}=\frac{\pi}{2}$
$\Rightarrow \text{I}=\frac{\pi}{4}$

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