Question:
Evaluate the following integrals: $\int \frac{2 x-1}{(x-1)^{2}} d x$
Solution:
Let $I=\int \frac{2 x-1}{(x-1)^{2}} d x$
Substituting $x-1=t \Rightarrow d x=d t$
$\Rightarrow I=\int \frac{2(t+1)-1}{t^{2}} d t$
$\Rightarrow I=\int \frac{2 t+1}{t^{2}} d t$
$\Rightarrow I=\int\left(\frac{2}{t}+\frac{1}{t^{2}}\right) d t$
$\Rightarrow I=2 \log |t|+\frac{1}{t}+c$
$\Rightarrow I=2 \log |x-1|+\frac{1}{x-1}+c$
Therefore, $\int \frac{2 x-1}{(x-1)^{2}} d x=2 \log |x-1|+\frac{1}{x-1}+c$
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