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Alpha uses the $!x$ notation for $$\frac{\Gamma(x+1,-1)}{e},$$ where $$\Gamma(x,a)=\int_x^{\infty}t^{-a}e^tdt,$$ the incomplete Gamma function.


The issue lay in the fact that the notation $!6$ means something very specific to Wolfram Alpha. You likely want it to mean $6$ factorial (i.e. $6\cdot 5\cdot 4\cdot 3 \cdot 2 \cdot 1$). However Wolfram Alpha interprets $!6$ to be the subfactorial of $6$ which is the number of permutations of $6$ elements which leaves no element fixed. In this case, $!6 = ...


Given that you tagged this with factorial I assume what you meant to write was: 6!/(3!*3!) You put the ! on the wrong side. n! is n factorial. !n is sub factorial. Next time just type in the words to make sure you have the right syntax(e.g., factorial(6)((factorial(6)*factorial(6)))


The integral expression that defines the log function is $$\log x=\int_1^x \frac{dt}{t}$$ for $x>0$. $\,$Obvioulsy, for $x<0$ we may write the equivalent forms $$\begin{align} \log (-x)&=\int_1^{-x} \frac{dt}{t}\\\\ &=\int_{-1}^{x} \frac{dt}{t}\\\\ &=-\int_{x}^{-1} \frac{dt}{t} \end{align}$$ As pointed out by Mathemagician1234, one ...


Notice all these solutions vary in very specific ways that are "isomorphic" under differentiation under the integral sign. For example, the Alpha and Online integrator solutions are scalar multiples of the arguments under composition of functions: $\frac{1}{6}\left( {\ln (x + 3) - \ln (3 - x)} \right) + C$= $\frac{1}{6}\left( {\ln \frac{(x + 3)}{(3 - x)}} ...

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