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whats the integral of 1/ ( x * (3-x) ) I know it should be a ln ( something)
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do u know hw to use partial fractions?
\[\frac{1}{x(3-x)} = \frac{A}{x}+\frac{B}{3-x}\]
\[1=A(3-x)+bx\]\[1=3A-Ax+Bx\]\[3A=1\]\[A=\frac{1}{3}\]\[-A+B=0\]\[\frac{-1}{3}+B=0\]so\[B=\frac{1}{3}\]
\[\frac{1}{x(3-x)}=\frac{1}{3x} + \frac{1}{3(3-x)}\]
Thanks i can integrate that.
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