\( \int_{-\infty}^\infty \frac{dx}{\left(4+x^2 \right) \sqrt{4+x^2}} \)
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\( \int_{-\infty}^\infty \frac{dx}{\left(4+x^2 \right) \sqrt{4+x^2}} \)
Evaluate the integral: $$\int_{-\infty}^\infty \frac{dx}{\left(4+x^2 \right) \sqrt{4+x^2}} $$
Imagination is much more important than knowledge.
Re: \( \int_{-\infty}^\infty \frac{dx}{\left(4+x^2 \right) \sqrt{4+x^2}} \)
Given \(\displaystyle \int_{-\infty}^{\infty}\frac{1}{(x^2+4)\sqrt{x^2+4}}dx = 2\int_{0}^{\infty}\frac{1}{(x^2+4)\sqrt{x^2+4}}dx\)
Now Let \(\displaystyle x= \frac{1}{t}\) and \(\displaystyle dx = -\frac{1}{t^2}dt\) and changing Limits ,
So integral convert into \(\displaystyle 2\int_{0}^{\infty}\frac{1}{(1+4t^2)^{\frac{3}{2}}}dt\)
Now let \(\displaystyle (4t^2+1) = u^2\) and \(\displaystyle tdt = \frac{u}{4}du\) and changing limit
we get \(\displaystyle 2\int_{1}^{\infty}\frac{u}{4}\cdot \frac{1}{u^3}du = \frac{1}{2}\int_{1}^{\infty}u^{-2}du = \frac{1}{2}\)
Now Let \(\displaystyle x= \frac{1}{t}\) and \(\displaystyle dx = -\frac{1}{t^2}dt\) and changing Limits ,
So integral convert into \(\displaystyle 2\int_{0}^{\infty}\frac{1}{(1+4t^2)^{\frac{3}{2}}}dt\)
Now let \(\displaystyle (4t^2+1) = u^2\) and \(\displaystyle tdt = \frac{u}{4}du\) and changing limit
we get \(\displaystyle 2\int_{1}^{\infty}\frac{u}{4}\cdot \frac{1}{u^3}du = \frac{1}{2}\int_{1}^{\infty}u^{-2}du = \frac{1}{2}\)
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