Maxima's answer might not be what you want, but it's correct. Where defined,
the functions log(x-a) and log(a-x) differ by a constant. Checking Maxima's answer,
we have:
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Maxima's answer might not be what you want, but it's correct. Where defined,
the functions log(x-a) and log(a-x) differ by a constant. Checking Maxima's answer,
we have:
(%i21) integrate(1/(a^2-x^2), x);
(%o21) log(x+a)/(2*a)-log(x-a)/(2*a)
(%i22) rat(diff(%,x));
(%o22) -1/(x^2-a^2)
Maybe you would like to use logabs : true
(%i23) integrate(1/(a^2-x^2), x), logabs;
(%o23) log(abs(x+a))/(2*a)-log(abs(x-a))/(2*a)
This Tracker item was closed automatically by the system. It was
previously set to a Pending status, and the original submitter
did not respond within 14 days (the time period specified by
the administrator of this Tracker).