# Special case of this general integral

We know that the integral of,

$$\int \frac{1}{x} dx = \ln x$$

ignoring any integration constants.

Consider the integral of some more `general' function,

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

i.e. in the $$y=z=0$$ case we recover the original integral ($$1/x$$)

Now, if I put the second integral into WolframAlpha, it spits out,

$$\int \frac{1}{\sqrt{x^2 + y^z +z^2}} dx = \ln \left(\sqrt{x^2+y^2+z^2} + x\right)$$.

I naively expected that in the $$y=z=0$$ case, this answer would revert to $$\ln x$$, but instead we get $$\ln 2x$$

What gives?

• Technically it's $\int\frac{1}{x}dx=\ln|x|+C$, where the locally constant function $C$ can have different values for the cases $x<0,\,x>0$.
– J.G.
Dec 4, 2018 at 18:30

## 1 Answer

Note that

$$\ln 2x=\ln x +\ln 2=\ln x +C$$

• Hmmm ok, but if I wanted to evaluate both over some limits $x=x_1 \rightarrow x_2$, it seems I would still get different answers? Dec 4, 2018 at 18:30
• @user1887919 No since the constant cancels out, let try also with some numerical example.
– user
Dec 4, 2018 at 18:31
• @user1887919 You are welcome! Refer also to that related OP, Bye
– user
Dec 4, 2018 at 18:36