<s>[INST] How do I solve differentil equation if the begining values are given? How do I solve differentil equation if the begining values are given?
$\frac{dx}{dt}=x^2+5x$ with x(0)=-3. I need to find x(t).
$\int \frac{dx}{x^2+5x}=\int dt$
So when I put it on Symbollab I get that left side is $-\frac{2}{5}arctanh(\frac{2}{5}(x+\frac{2}{5}))$, I already see that is going to be hard to extract x from here.
What have I done?
$\int \frac{dx}{x^2+5x}=\int \frac{dx}{x^2(1+\frac{x}{5})}$ than I use substitution that $1+\frac{x}{5}$ is u, than $-\frac{x^2}{5}du$.
Than I get $\frac{-1}{5}ln(1+\frac{x}{5})=t+C$.
$ln(1+\frac{x}{5})=-5t+C$
$1+\frac{x}{5}=e^{-5t}C$
$\frac{x}{5}=e^{-5t}C-1$
$x=\frac{5}{e^{-5t}C-1}$
$-3=\frac{5}{e^{0}C-1}$
$-3=\frac{5}{C-1}$
Than I calculate C. This all seems nice to me.
But when I put than integral in WolframAlpha I get that $\int \frac{dx}{x^2+5x}=\frac{1}{5}(log(x)-log(x+5))+constant$
Why I don't get when I integrate those results from WolframAlpha or Symbollab?
 [/INST] If $y = \mbox{arc}\tanh x$, then $$x = \tanh y =\frac{e^y - e^{-y}}{e^y+e^{-y}}= \frac{e^{2y} - 1}{e^{2y}+1}.$$  So
$$ x(e^{2y}+1) = e^{2y}-1$$
or $$e^{2y}(x-1) = -x-1$$
So $$e^{2y} = \frac{1+x}{1-x}.$$  That is
$$2y = \log\left( \frac{1+x}{1-x}\right)$$
and
$$y= \frac{1}{2}(\log(1+x)-\log(1-x)).$$
So we have $\mbox{arc}\tanh x = \frac{1}{2}(\log(1+x)-\log(1-x).$$
Partial fractions is giving the $\log$ version of the answer and WA is giving the $\mbox{arc}\tanh$ version.
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