Question

Evaluate : $\int \frac{1}{1-4x}dx$

Answer 1

Suppose that

$I=\int \frac{1}{1-4x}dx$

By putting $1-4x=t$

$\therefore -4dx=dt$

$\Rightarrow dx=-\frac{1}{4}dt$

$\therefore I=-\frac{1}{4}\int \frac{1}{t}dt$

$=-\frac{1}{4}{\log }_{e}t$

$\Rightarrow I=-\frac{1}{4}{\log }_e(1-4x)$.