_0^1 dx [ax + b(1 - x)] ^2 = - Vidyadip

MCQ

$\int_0^1 {\frac{{dx}}{{{{[ax + b(1 - x)]}^2}}}} = $

A
$\frac{a}{b}$
B
$\frac{b}{a}$
C
$a\,b$
✓
$\frac{1}{{a\,b}}$

✓

Answer

Correct option: D.

$\frac{1}{{a\,b}}$

d
(d) Let $I = \int_0^1 {\frac{{dx}}{{{{[(a - b)x + b]}^2}}}} $

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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