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

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(D) Let $I = \int_0^1 \frac{dx}{[(a - b)x + b]^2}$.Substitute $t = (a - b)x + b$,then $dt = (a - b)dx$,which implies $dx = \frac{dt}{a - b}$.When $x =

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$\frac{1}{ab}$

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(D) Let $I = \int_0^1 \frac{dx}{[(a - b)x + b]^2}$.Substitute $t = (a - b)x + b$,then $dt = (a - b)dx$,which implies $dx = \frac{dt}{a - b}$.When $x = 0$,$t = b$. When $x = 1$,$t = a$.Substituting these into the integral:$I = \int_b^a \frac{1}{t^2} \cdot \frac{dt}{a - b} = \frac{1}{a - b} \int_b^a t^{-2} dt$.Evaluating the integral:$I = \frac{1}{a - b} \left[ -\frac{1}{t} \right]_b^a = \frac{1}{a - b} \left( -\frac{1}{a} - (-\frac{1}{b}) \right)$.$I = \frac{1}{a - b} \left( \frac{1}{b} - \frac{1}{a} \right) = \frac{1}{a - b} \left( \frac{a - b}{ab} \right) = \frac{1}{ab}$.

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

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