MCQ
$\int_0^1 \frac{d x}{[a x+b(1-x)]^2}=$
  • A
    $\frac{ a }{ b }$
  • B
    $\frac{ b }{ a }$
  • C
    ab
  • $\frac{1}{a b}$

Answer

Correct option: D.
$\frac{1}{a b}$
(D)
Let $I =\int_0^1 \frac{1}{[ a x+ b (1-x)]^2} d x$
$=\int_0^1 \frac{1}{[( a - b ) x+ b ]^2} d x$
Put $( a - b ) x+ b = t \Rightarrow( a - b ) d x= dt$
When $x=0, t = b$ and when $x=1, t = a$
$\therefore \quad I=\frac{1}{a-b} \int_b^a \frac{1}{t^2} d t$
$\begin{array}{l}=\frac{1}{(a-b)}\left[-\frac{1}{t}\right]_b^a \\ =\frac{1}{(a-b)}\left(\frac{a-b}{a b}\right)\end{array}$
$\therefore \quad I=\frac{1}{a b}$

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