# Tag Info

18

Let $u=x^2, du=2xdx$ Then the Integral becomes $$\int_{0}^{\infty}\frac{1}{u^4+2u^2+1}du$$ Now notice $u^4+2u^2+1=(u^2+1)^2$, so the integral becomes: $$\int_{0}^{\infty}\frac{1}{(u^2+1)^2}du$$ Let $u=\tan(z),du=\sec^2(z)dz$ So the integral becomes : ...

11

Your conjecture is a trivial consequence of Frullani's theorem after the substitution $x=e^{-z}$.

9

Assume $x>1$ and write $${1\over\sqrt{x^2-1}}={1\over\sqrt{x^2-1}}\>{\sqrt{x^2-1}+x\over \sqrt{x^2-1}+x}={1+{x\over\sqrt{x^2-1}}\over x+\sqrt{x^2-1}}={u'(x)\over u(x)}$$ with $u(x):=x+\sqrt{x^2-1}>0$. It follows that $$\int{dx\over\sqrt{x^2-1}}=\log\bigl(u(x)\bigr)+C=\log\bigl(x+\sqrt{x^2-1}\bigr)+C\ .$$

7

Try integration by parts, differentiating $z$ and integrating $ze^{-\frac{z^2}{2}}$. You'll need the fact that $$\int_{-\infty}^{\infty}e^{-\frac{z^2}{2}}\;dz=\sqrt{2\pi}$$

6

Let the function $I(a)$ be the integral given by \begin{align} I(a)&=\int_{-\infty}^\infty e^{-az^2}\,dz\\\\ &=\sqrt{\frac{\pi}{a}} \end{align} Then, note that $I'(1/2)$ is \begin{align} I'(1/2)&=-\int_{-\infty}^\infty z^2e^{-\frac12 z^2}\,dz\\\\ &=-\sqrt{2\pi} \end{align} Therefore, we find $$\int_{-\infty}^\infty ... 6 By subbing x=\sqrt{z}, then \frac{1}{z^2+1}=u$$\int_{0}^{+\infty}\frac{2x}{(x^4+1)^2}\,dx = \int_{0}^{+\infty}\frac{dz}{(z^2+1)^2}=\frac{1}{2}\int_{0}^{1}u^{1/2}(1-u)^{-1/2}\,du$$that by Euler's beta function and \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi} equals:$$ \frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)}{2\,\Gamma(2)} = ...

6


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