# Computing $\int\frac {du}{\sqrt{u^2 + s^2}} =\log \lvert (u + \sqrt{u^2 + s^2}) \rvert$ with a substitution

Can someone please show me where I am going wrong? It seems there is a contradiction in the formula for the the following integral.

$$\int\frac {du}{\sqrt{u^2 + s^2}} = \log \bigl\lvert u + \sqrt{u^2 + s^2}\bigr\rvert$$

Now, can't we rewrite the integral as follows using the substitution z = -u?

\begin{align} \int\frac {du}{\sqrt{u^2 + s^2}} &= \int\frac {du}{\sqrt{(-u)^2 + s^2}} = \int\frac {-dz}{\sqrt{z^2 + s^2}}\\ & = -\log \bigl\lvert z + \sqrt{z^2 + s^2}\bigr\rvert = -\log \bigl\lvert (-u + \sqrt{u^2 + s^2})\bigr\rvert \end{align} But, clearly $$\;\log \bigl\lvert u + \sqrt{u^2 + s^2}\bigr\rvert \ne -\log \bigl\lvert -u + \sqrt{u^2 + s^2} \bigr\rvert$$

Also, the quantity inside $$\log \bigl\lvert u + \sqrt{u^2 + s^2})\bigr\rvert$$ is clearly non-negative for any choice of $$u$$ and $$s$$, so why do we need the absolute value of the quantity in our integral?

## 2 Answers

Both are not equal . But you'll get a term which will get added to the arbitrary constant and get cancelled in definite integral.

$$I=-\log|-u+\sqrt{u^2+s^2}|+c = \log\bigg\vert\frac{1}{-u+\sqrt{u^2+s^2}}\bigg\vert+c$$

$$I = \log\bigg\vert\frac{u+\sqrt{u^2+s^2}}{-u^2+u^2+s^2}\bigg\vert+c=\log\vert u+\sqrt{u^2+s^2}\vert-\log s^2+c = \log\vert u+\sqrt{u^2+s^2}\vert +k$$

where $$k=c-\log s^2$$

And this gets cancelled in a definite integral, so both are equivalent.

You have a faster way, using some hyperbolic trigonometry:

Set $$t=\dfrac us$$, so that $$\mathrm du=s\,\mathrm dt$$. The integral becomes, supposing $$s>0\,$$: $$\int\frac {\mathrm du}{\sqrt{u^2 + s^2}}=\int\frac {\not\! s\,\mathrm dt}{\not \!s\sqrt{ t^2 + 1}}=\int\frac {\mathrm dt}{\sqrt{ t^2 + 1}}=\operatorname{argsinh}t$$ and it is known that $$\operatorname{argsinh}t =\ln\bigl(t+\sqrt{t^2+1}\bigr).$$

• Nice to see this notation with arg ! It remembers to me the old time. Cheers. – Claude Leibovici Aug 14 at 6:33
• @ClaudeLeibovici: Actually, I should have written $\operatorname{argsh}$, but I didn't dare ;o) – Bernard Aug 14 at 8:35