Answer
$5\ln|x-2|-\frac{8}{x-2}+C$
Work Step by Step
Let $\frac{5x-2}{(x-2)^{2}}=\frac{A}{x-2}+\frac{B}{(x-2)^{2}}$
$⇒ 5x-2= A(x-2)+B$
Equating the coefficient of x and constant term, we get
A=5, -2A+B= -2
⇒B= 8
Thus the integrand is given by
$\frac{5x-2}{(x-2)^{2}}=\frac{5}{x-2}+\frac{8}{(x-2)^{2}}$
Therefore, $\int \frac{5x-2}{(x-2)^{2}}dx= 5\int\frac{1}{x-2}dx+8\int\frac{1}{(x-2)^{2}}dx$
$= 5 \ln |x-2|+ 8\times(\frac{-1}{x-2})+C$
$=5\ln|x-2|-\frac{8}{x-2}+C$
