Question

${\int }_0^1\frac{dx}{[ax+b(1-x){]}^2}=$

• A. $\frac{a}{b}$
• B. $\frac{b}{a}$
• C. $ab$
• D. $\frac{1}{ab}$

Answer 1

The correct option is D $\frac{1}{ab}$

$LetI={\int }_0^1\frac{dx}{\left[\right(a-b)x+b{]}^2}$
$Putt=(a-b)x+b\Rightarrow dt=(a-b)dx$
$Asx=1\Rightarrow t=aandx=0\Rightarrow t=b,then$
$I=\frac{1}{a-b}{\int }_b^a\frac{1}{t^2}dt=\frac{1}{(a-b)}{[-\frac{1}{t}]}_b^a=\frac{1}{(a-b)}\left(\frac{a-b}{ab}\right)=\frac{1}{ab}$