How would you integrate $\sqrt{1+\frac{1}{x^2}}$ [duplicate]

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I need to integrate $\sqrt{1+\frac{1}{x^2}}$

I've tried to let $u=1/x^2$ but end up with, $\int \frac{\sqrt{1+u^2}}{u^{3/2}}du$ . I attempted to then substitute $u=\tan\theta$ and lead me to $-\frac{1}{2}\int \frac{1}{(\sin\theta\cos\theta)^{3/2}}d\theta$.

Not sure how to progress from here. Is there any alternative way of doing this?

marked as duplicate by Chris Culter, Claude Leibovici, Najib Idrissi, Martin Sleziak, kingW3Feb 14 '15 at 10:13

$$\int\sqrt{1+\frac{1}{x^2}}\,dx=\int \frac{\sqrt{1+x^2}}{x}\,dx=\int \frac{x\sqrt{1+x^2}}{x^2}\,dx$$ Let $1+x^2=u^2$ so that$$\int \frac{x\sqrt{1+x^2}}{x^2}\,dx=\int \frac{u^2}{u^2-1}\,du$$ hence we have that $$\int \frac{u^2}{u^2-1}\,du=u+\frac{1}{2}\ln |\frac{u-1}{u+1}|+c$$

user62498 gave you the good answer but, if I may, let me detail a bit more $$\frac{u^2}{u^2-1}=\frac{u^2-1+1}{u^2-1}=1+\frac{1}{u^2-1}$$ Now use partial fraction decomposition and show that $$\frac{1}{u^2-1}=\frac{1}{2 (u-1)}-\frac{1}{2 (u+1)}$$ Now, you have the the small pieces to make the integration easy.

Let $x=\tan y,y=\arctan x\implies-\dfrac\pi2\le y\le\dfrac\pi2$

$$\implies 1+\frac1{x^2}=\frac1{\sin^2y},dx=\frac{dy}{\cos^2y}$$

$$\int\sqrt{1+\frac1{x^2}}dx=\int\frac{dy}{|\sin y|\cos^2y}$$

$$=\text{sign of}(\sin y)\int\frac{\sin y\ dy}{(1-\cos^2y)\cos^2y}$$

Setting $\sin y=u,$

$$\int\frac{\sin y\ dy}{(1-\cos^2y)\cos^2y}=\int\frac{du}{u^2(1-u^2)}=\int\frac{du}{1-u^2}+\int\frac{du}{u^2}$$

Hope you can take it from here

Let $x=\cot\theta$, so $dx=-\csc^{2}\theta$ and $\displaystyle\int\sqrt{1+\frac{1}{x^2}} \;dx=\int\sec\theta \;(-\csc^{2}\theta) \;d\theta$.

Now let $u=\sec\theta$, $dv=-\csc^{2}\theta d\theta$ to get

$\int\sec\theta \;(-\csc^{2}\theta) \;d\theta=\sec\theta\cot\theta-\int\sec\theta\;d\theta=\csc\theta-\ln\lvert\sec\theta+\tan\theta\rvert+C$

$\displaystyle=\sqrt{x^2+1}-\ln\left|\frac{\sqrt{x^2+1}+1}{x}\right|+C$.

• This answer seems a bit off from the previous answers but +1 for the method regardless. – sachinruk Sep 17 '14 at 0:30
• Thanks - my answer does look different, but I checked the derivative before posting it. – user84413 Sep 17 '14 at 0:34