Evaluating $\int{\frac{1}{\sqrt{x^2+y^2}}\mathrm dx}$

Question

Attempting to calculate $\displaystyle \int{\dfrac{1}{\sqrt{x^2+y^2}}\mathrm dx}$, $$\int{\dfrac{1}{\sqrt{x^2+y^2}}\mathrm dx}=\int{\frac{1}{\sqrt{(y\tan\theta)^2+y^2}}y\sec^2\theta \mathrm d\theta}=\int{\sec\theta d\theta}=\ln(\sec\theta +\tan\theta)=\ln\left(\sqrt{\left(\frac{x}{y}\right)^2+1}+\frac{x}{y}\right)=\ln\left(\frac{1}{x}\left(\sqrt{x^2+y^2}+x\right)\right),$$ where $x=y\tan\theta$

However Wolfram Integrator somehow returns $$\ln\left(2\left(\sqrt{x^2+y^2}+x\right)\right)$$ as the answer. Where did I go wrong? Many thanks.

Answer 1

To find out the reason you only requires some basic knowledge of logarithms. In short, there is only a constant difference between these two logarithms.

Since we are integrating with respect to $x$, we regard $y$ as constant and therefore $\ln y$ is const. Let's add the constant term of indefinite integral: \begin{align} \int\frac1{\sqrt{x^2+y^2}}dx & = \ln\left(\frac1y\left(\sqrt{x^2+y^2}+x\right)\right)+C \\ & = \ln\left(\sqrt{x^2+y^2}+x\right)-\ln y+C \\ & = \ln\left(\sqrt{x^2+y^2}+x\right)+C_0 \\ & = \ln\left(\sqrt{x^2+y^2}+x\right)+\ln 2+C' \\ & = \ln\left(2\left(\sqrt{x^2+y^2}+x\right)\right)+C' \\ \end{align} so the two answers are exactly the same.

Answer 2

Hint: after you fix the mistake in the last step, differentiate the function $$x\mapsto \ln\left(\frac{1}{y}\left(\sqrt{x^2+y^2}+x\right)\right)-\ln\left(2\left(\sqrt{x^2+y^2}+x\right)\right)$$ with respect to $x$. It's easy to do so in your head. Also do not forget that you should consider the absolute value appropriately ($\int \frac 1x\mathrm dx)=\ln (|x|)$) and to add an arbitrary constant when you integrate.

Answer 3

One alternative approach via $\sin(ix)=\sinh(x)$ or $\sin^{-1}(ix)=i\sinh^{-1}(x)$ and the known the formula $$ \int{\dfrac{1}{\sqrt{a^2-x^2}}\mathrm dx}=\sin^{-1}\frac{x}{a} $$ then if let $x=it$ $$ \int{\dfrac{1}{\sqrt{x^2+y^2}}\mathrm dx}=i\int{\dfrac{1}{\sqrt{y^2-t^2}}\mathrm dt}=i\sin^{-1}\frac{t}{y}=i\sin^{-1}\left(-i\frac{x}{y}\right)=-i^2\arcsin{\frac{x}{y}}=\arcsin{\frac{x}{y}} $$ where the oddity of $\sin^{-1}(x)$ has been used as well.