Question
$\int\frac{2\text{x}-1}{(\text{x}-1)^2}\text{ dx}$

Answer

$\int\Big[\frac{2\text{x}-1}{(\text{x}-1)^2}\text{ dx}\Big]$
Let $\text{x}-1=\text{t}$
$\Rightarrow\text{x}=1+\text{t}$
$\Rightarrow1=\frac{\text{dt}}{\text{dx}}$
Now, $\int\Big[\frac{2\text{x}-1}{(\text{x}-1)^2}\text{ dx}\Big]$
$=\int\Big[\frac{2(\text{t}+1)-\text{t}}{\text{t}^2}\Big]\text{dt}$
$=\int\Big(\frac{2\text{t}+1}{\text{t}^2}\Big)\text{dt}$
$=2\int\frac{\text{dt}}{\text{t}}+\int\text{t}^{-2}\text{dt}$
$=2\log|\text{t}|+\frac{\text{t}^{-2+1}}{-2+1}+\text{C}$
$=2\log(\text{x}-1)-\frac{1}{\text{x}-1}+\text{C}$

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