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