Trouble finding $\int \frac{\sqrt{1+x^2}}{x}$

Question

I have tried the substitution $x=\sinh u$. This got me to $\,\int \frac{\cosh^2u}{\sinh u}du$, and through an identity I got $\int(\operatorname{csch} u+\sinh u)du$. But integrating this, and substituting $x$ back does not get me the correct answer. May someone tell me where I've gone wrong.

Answer 1

Hint

Even if $x=\sinh u$ does not look to be the best possible substitution, you can continue. $$I=\,\int \frac{\cosh^2(u)}{\sinh(u)}du=\,\int \frac{1+\sinh^2(u)}{\sinh(u)}du=\,\int \sinh(u)du+\,\int \frac{1}{\sinh(u)}du$$ Now, consider the last integrand$$\frac{1}{\sinh(u)}=\frac{\cosh^2(\frac u2)-\sinh^2(\frac u2)}{2\sinh(\frac u2)\cosh(\frac u2)}=\frac 12\Big(\frac{\cosh(\frac u2)}{\sinh(\frac u2)}-\frac{\sinh(\frac u2)}{\cosh(\frac u2)}\Big)$$ So, $$I=\,\int \sinh(u)du+\frac 12 \int \frac{\cosh(\frac u2)}{\sinh(\frac u2)}du-\frac 12 \int \frac{\sinh(\frac u2)}{\cosh(\frac u2)} du$$ For the last integrals make another change of variable $u=2v$ and arrive to $$I=\,\int \sinh(u)du+\int \frac{\cosh(v)}{\sinh(v)}dv- \int \frac{\sinh(v)}{\cosh(v)} dv$$

I am sure that you can take from here.

Answer 2

Where things went wrong, if they did, cannot be determined without more information. What is written down so far is correct.

May I suggest a simpler way? Our function is $\frac{x\sqrt{1+x^2}}{x^2}$. Let $u^2=1+x^2$. Then $x\,dx=u\,du$, and we are integrating $\frac{u^2}{u^2-1}$.

Answer 3

Hint:

Let $x = \tan{u} \implies dx = \sec^2{u} \ du$. Then,

$$\int \frac{\sqrt{ 1 + x^2}}{x}dx = \int\frac{du}{\sin{u}\cos^2u} = \int \csc {u} \sec^2{u} \ du = \int \csc \theta(1 + \tan^2\theta) \ du = \int \csc\theta \ du + \int \csc{u}\tan^2u \ du = \int \csc{u} \ du + \int \dfrac{\sin{u}}{\cos^2u} \ du.$$

It is essentially a simple rearrangement and substitution for different concepts to make this problem look a lot simpler than it seems.

Answer 4

A simpler substitution, which is similar to a secant substitution, can be used for this integral. For positive $x$, let $x=\operatorname{csch}{\theta}$ where $\theta>0$, then $dx=-\operatorname{csch}{\theta}\coth{\theta}\,d\theta$ and $$\sqrt{1+x^2}=\sqrt{1+\operatorname{csch}^2{\theta}}=\sqrt{\coth^2{\theta}}=\left|\coth{\theta}\right|=\coth{\theta}$$ Then $$\int{\frac{\sqrt{1+x^2}}{x}dx}=-\int{\frac{\operatorname{csch}{\theta}\coth^2{\theta}}{\operatorname{csch}{\theta}}d\theta}=-\int{\coth^2{\theta}\,d\theta}$$ The resulting integral can be evaluated using the identity $\coth^2{\theta}=1+\operatorname{csch}^2{\theta}$ $$-\int{\coth^2{\theta}\,d\theta}=-\int{\left(1+\operatorname{csch}^2{\theta}\right)\,d\theta}=\coth{\theta}-\theta+C$$ Using the identities $\sin{i\theta}=i\sinh{\theta}$ and $\cos{i\theta}=\cosh{\theta}$, a reference triangle can be made to undo the substitution.

Reading from the triangle, we have $\cot{i\theta}=\frac{1}{i}\sqrt{1+x^2}=\frac{1}{i}\coth{\theta}$, therefore $$\sqrt{1+x^2}-\operatorname{arcsch}{x}+C\qquad x>0$$ Similarly for negative $x$, let $x=-\operatorname{csch}{\theta}$, where $\theta>0$, and $dx=\operatorname{csch}{\theta}\coth{\theta}\,d\theta$ and follow the above steps, obtaining $$\sqrt{1+x^2}-\operatorname{arcsch}{\left(-x\right)}+C\qquad x<0$$ Combining the two branches gives $$\sqrt{1+x^2}-\operatorname{arcsch}{\left|x\right|}+C\qquad x\neq0$$