I know that
$$\int \frac{x-a}{x(1-x)}dx = -[a\ln x+ (1-a) \ln(1-x)]+const$$
but how can I prove it?
Let $$\frac{x-a}{x(1-x)}=\frac Ax + \frac B{1-x}$$
At $x=0$, $A = \frac{-a}{1} = -a$
At $x=1$, $B = \frac{1-a}{1} = 1-a$
So we have $$\frac{-a}{x} + \frac{1-a}{1-x}$$
Could you proceed further?
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Let $$\frac{x-a}{x(1-x)}=\frac Ax + \frac B{1-x}$$
At $x=0$, $A = \frac{-a}{1} = -a$
At $x=1$, $B = \frac{1-a}{1} = 1-a$
So we have $$\frac{-a}{x} + \frac{1-a}{1-x}$$
Could you proceed further?