Question

$\underset{0}{\overset{1}{\int }}\frac{dx}{[ax+b(1-x){]}^2}$ is equal to

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

Answer 1

Answer: (B)

$\underset{0}{\overset{1}{\int }}\frac{dx}{{\left[ax+b\left(1-x\right)\right]}^2}=\underset{0}{\overset{{}^1}{\int }}{\left[b+\left(a-b\right)x\right]}^{-2}dx$
$={\left|\frac{b+\left(a-b\right)x^{-1}}{-\left(a-b\right)}\right|}_0^1$
$=-\frac{1}{a-b}{\left([b+\left(a-b\right)\right]}^{-1}-{\left[b\right]}^{-1}])$
$=-\frac{1}{a-b}\left[\frac{1}{a}-\frac{1}{b}\right]$
$=-\frac{1}{a-b}\cdot \frac{b-a}{ab}=\frac{1}{ab}$