Question

Evaluate: $\frac{1}{3}\int \frac{1}{(x-1{)}^2}dx$

Answer 1

$I=\frac{1}{3}\int \frac{1}{(x-1{)}^2}dx$

Let $x-1=t$

$dx=dt$

$I=\frac{1}{3}\int \frac{dt}{t^2}$

$I=\frac{1}{3}\int t^{-2}dt$

$I=\frac{1}{3}(-t^{-1})+C$

$I=-\frac{1}{3}\times \frac{1}{x-1}+C$

$I=-\frac{1}{3(x-1)}+C$