Question

calculate the integral x/[(x-a)(x-b)]
when a=b

Answer 1

With a=b we can write\[\int\frac x{(x-a)^2}dx\]now a little u substitution\[u=x-a\to x=u+a\to du=dx\]\[\int\frac{u+a}{u^2}du=\int u^{-1}+au^{-2}du\]Got it from here?

Answer 2

but shouldnt we split it into partial fractions first??

Answer 3

I don't see a need to...

Answer 4

but sure, that should give the same result
seems like an unnecessary pain though

Answer 5

but i think that is what they want here and that is where i am lost

Answer 6

\[\int\frac x{(x-a)^2}=\int\frac A{x-a}+\frac B{(x-a)^2}dx\]\[A(x-a)+B=x\to Ax+B-Aa=x\]so we get\[A=1\]\[B-a=0\to B=a\]so our integral is\[\int\frac1{x-a}+\frac a{(x-a)^2}dx\]looks to me like we're gonna get the same answer both ways, which is a good thing :D

Answer 7

thank you