Essentially this is the substitution rule. More formally, the substitution $V=v(x')$ yields
$$\int_0^{x}\frac{dv/dx'}{\sqrt{1+v(x')^2}}\,dx'=\int_0^{v(x)}\frac{1}{\sqrt{1+V^2}}dV$$
where I've switched to definite integration to make sure the changes of variable are all clear. Since $$\frac{d}{dV}\sinh^{-1} V=\frac{1}{\sqrt{1+V^2}},$$ this integrates to $ \Big[\sinh^{-1}V\Big]_{V=0}^{v(x)}=\sinh^{-1} v(x).$ In solutions, the above definite is instead typically presented as the indefinite integral
$$\int \frac{dv/dx}{\sqrt{1+v(x)^2}}dx=\int\frac{dv}{\sqrt{1+v^2}}=\sinh^{-1}v(x)+C$$ which is what WolframAlpha has done.
WolframAlpha is just some natural language input parsing on top of a limited form of Wolfram Language: the programming language used in Mathematica.
If Mathematica were given {{1,-1},{0,2}}^r
, it would not take the $r^{\text{th}}$ power of the $2 \times 2$ matrix in the usual linear algebra sense; it would instead take the $r^{\text{th}}$ power of each entry. You would have to write MatrixPower[{{1,-1},{0,2}},r]
for the desired result.
WolframAlpha is a little bit better at guesing what you want: if you write {{1,-1},{0,2}}^r
, it seems to guess that you mean a matrix power. If you write ({{1,-1},{0,2}})^(r)
, that's too confusing for WolframAlpha, and it falls back to the default (and probably wrong) interpretation.
So that's the difference between the two expressions. In the first case, WolframAlpha is taking matrix powers and adding them up. In the second case, WolframAlpha is dealing with each entry of the matrix separately.
If you do not want to rely on WolframAlpha guessing what you want, you can use syntax closer to Mathematica syntax to disambiguate. For example, you can ask WolframAlpha to compute
sum of MatrixPower[{{1,-1},{0,2}},r] from r=1 to n
and when you get the output (which is the same as your first result), you can be certain that WolframAlpha is taking the correct kind of matrix power.
Best Answer
Wolfram Alpha depends on the Wolfram Language in Mathematica and it takes some getting used to the non intuitive way that Mathematica works.
Your example query
to WA was interpreted as the Wolfram Language command
This returns essentially five different solutions. The first three are $$ L=0 \tag{1}$$ $$ C=0,\quad D=0 \tag{2}$$ $$ C\ne0,\quad D\ne0,\quad D=C,\quad L=0 \tag{3}$$ The next two depend on an integer $\,n\in\mathbb{Z}\,$ as follows $$ \sqrt{L}\ne0,\quad m=\frac{in}{\sqrt{L}} \tag{4}$$ $$ D\!\ne\!0, C\!\ne\!0, \sqrt{L}\!\ne\!0, t\!=\!-m\pi \!+\! \frac{i\pi n}{\sqrt{L}} \!+\!\frac{\log(\frac CD)}{2\sqrt{L}} \tag{5}$$ Note that for equation $(3)$ the $\,D\ne0\,$ is replaced with $\,C\ne0\,$ in the WA display.
So far so good. Now you appended "$\,\texttt{, solve for L}\,$" to your query and WA interpreted this more literally as
Note the replacement of
Solve
forReduce
and the list of variables is only{L}
. The two solutions returned which depend on an integer $\,n\in\mathbb{Z}\,$ are $$ L = -\frac{n^2}{m^2} \tag{6} $$ $$ L = \frac{(2i\pi n+\log(\frac CD))^2}{4(m\pi+t)^2} \tag{7} $$ As you noted, the $\,n^2\,$ is dropped from the equation $(6)$ for some reason. Similarly, the $\,2i\pi n\,$ is dropped in equation $(7)$ for some reason. Probably a bug in the WA result display. The underlying Wolfram Language code returns the correct results, but WA mangles them.You asked
It is very hard to answer this because the WA system is proprietary and Wolfram does not give enough information about how to control what WA does interfacing to Mathematica. As I explained, in the first query, WA decided to use
Reduce
and in the second query,Solve
was used. Hard to explain why it does this.Of course, the best way is to use Mathematica's Wolfram Language directly which is possible using the Wolfram Cloud.