MCQ
$\int_0^1 {\frac{{dx}}{{{{[ax + b(1 - x)]}^2}}}} = $
- A$\frac{a}{b}$
- B$\frac{b}{a}$
- C$a\,b$
- ✓$\frac{1}{{a\,b}}$
Put $t = (a - b)x + b \Rightarrow dt = (a - b)dx$
As $x = 1 \Rightarrow t = a$ and $x = 0 \Rightarrow t = b$, then
$I = \frac{1}{{a - b}}\int_b^a {\frac{1}{{{t^2}}}} dt = \frac{1}{{(a - b)}}\left[ { - \frac{1}{t}} \right]_b^a $
$= \frac{1}{{(a - b)}}\left( {\frac{{a - b}}{{ab}}} \right) = \frac{1}{{ab}}$.
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