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integrate 1/((x-2)(x+3)
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partial fractions for this one do you know how to do it?
\[\frac{1}{(x-2)(x+3)}=\frac{a}{x-2}+\frac{b}{x+3}\] solve for "a" and "b"
start with \[a(x+3)+b(x-2)=1\] replace x by 2 get \[5a=1\implies a=\frac{1}{5}\] and then replace x by -3 get \[-5b=1\implies b=-\frac{1}{5}\] then rewrite and compute the two integrals, which will both be logs
you can just about do it in your head, you will get \[\frac{1}{5}\ln(x-2)-\frac{1}{5}\ln(x+3)\]
thank u, u just made this so clear to me!
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