what is the integral $\int_{0}^1 \sqrt{(1+(1/x^2))}dx$

Question

Can this integral can be evaluated? When I attempted it I got an answer of infinity. Maybe I don't know the correct method of solving this integration problem. Please provide the correct answer with an explanation. Thanks

Answer 1

We can use comparison to evaluate this improper integral:

$\sqrt{1+1/x^2}\geq\sqrt{1/x^2}=1/x$ for $x>0$.

Thus we have that $\int_a^1 \sqrt{1+1/x^2} dx\geq \int_a^1 \frac{1}{x}dx$ for all $a\in(0,1)$. But $\lim_{a\rightarrow 0^+}\int_a^1 \frac{1}{x}dx=+\infty$, so $\lim_{a\rightarrow 0^+}\int_a^1 \sqrt{1+1/x^2} dx=+\infty$ as well.

Answer 2

$$\int_0^1 \left(\sqrt{1+\frac{1}{x^2}}\right)dx=$$ $$\int_0^1 \left(\frac{\sqrt{x^2+1}}{x}\right)dx=$$

(Substitute $x=\tan(u)$ and $dx=\sec^2(u)du$. Then $\sqrt{x^2+1}=\sqrt{\tan^2(u)+1}=\sec(u)$ and $u=\tan^{-1}(x)$):

$$\int_0^1 \left(\csc(u)\sec^2(u)\right)du=$$ $$\int_0^1 \left((\tan^2(u)+1)\csc(u)\right)du=$$ $$\int_0^1 \left(\csc(u)+\tan(u)\sec(u)\right)du=$$ $$\int_0^1\left(\tan(u)\sec(u)\right)du+\int_0^1\left(\csc(u)\right)du=$$ $$\int_0^1\left(\frac{\sin(u)}{\cos^2(u)}\right)du+\int_0^1\left(\csc(u)\right)du=$$

(Substitute $s=\cos(u)$ and $ds=-\sin(u)du$):

$$\int_0^1\left(-\frac{1}{s^2}\right)ds+\int_0^1\left(\csc(u)\right)du=$$ $$-\int_0^1\left(\frac{1}{s^2}\right)ds+\int_0^1\left(\csc(u)\right)du=$$ $$-\left[-\frac{1}{s}\right]_{0}^{1}+\int_0^1\left(\csc(u)\right)du=$$ $$\left[\frac{1}{s}\right]_{0}^{1}+\left[-\ln\left(\cot(u)+\csc(u)\right)\right]_{0}^{1}=$$ $$\left[\frac{1}{s}\right]_{0}^{1}-\left[\ln\left(\cot(u)+\csc(u)\right)\right]_{0}^{1}=$$ $$\left[\frac{1}{\cos(u)}\right]_{0}^{1}-\left[\ln\left(\cot(u)+\csc(u)\right)\right]_{0}^{1}=$$

$$\left[\frac{1}{\cos(\tan^{-1}(x))}\right]_{0}^{1}-\left[\ln\left(\cot(\tan^{-1}(x))+\csc(\tan^{-1}(x))\right)\right]_{0}^{1}=$$

$$\left(\sqrt{2}-1\right)-\left(\ln(1+\sqrt{2})-intermedate\right)=$$ $$intermedate+\sqrt{2}-1-\sinh^{-1}(1)$$

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Putting zero in this equasion gives u an intermedate $\ln\left(\cot(\tan^{-1}(x))+\csc(\tan^{-1}(x))\right)$

Answer 3

Let $$I=\int \sqrt{1+\frac{1}{x^2}}.dx $$ $$= \int \sqrt{\frac{x^2+1}{x^2}}dx$$ $$=\int \frac{\sqrt{x^2+1}}{x}dx$$ Make the substitution Put $x=\tan\theta \implies dx=\sec^2\theta d\theta$ $$=\int \frac{\sqrt{\tan^2\theta+1}}{\tan\theta}sec^2\theta d\theta$$ $$=\int \frac{\sec\theta}{tan\theta}\sec^2\theta d\theta$$ $$= \int \frac{\cos\theta}{\sin\theta\cos\theta}\sec^2\theta d\theta$$ $$= \int \operatorname{cosec}\theta\sec^2\theta.d\theta$$ Integrating by parts $$=\operatorname{cosec}\theta\tan\theta-\int(-\operatorname{cosec}\theta cot\theta)\tan\theta d\theta$$ $$=\operatorname{cosec}\theta\tan\theta+\int \operatorname{cosec}\theta d\theta$$ $$=\operatorname{cosec}\theta\tan\theta+\ln |\operatorname{cosec}\theta - \cot\theta| $$ As $\tan\theta = x \implies \cot\theta =\frac{1}{x}$

And $\operatorname{cosec}\theta=\sqrt{\cot^2\theta+1} \implies \operatorname{cosec}\theta=\sqrt{\frac{1}{x^2}+1} $

$\implies \operatorname{cosec}\theta = \frac{\sqrt{x^2+1}}{x}$

So we have $$I=\frac{\sqrt{x^2+1}}{x}.x+\ln\left|\frac{\sqrt{x^2+1}}{x}-\frac1x\right|$$ $$I=\sqrt{x^2+1}+\ln\left|\frac{\sqrt{x^2+1}-1}{x}\right|$$