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  4. ∫ limits0 1(dx/[ax+b(1-x)]2) is equal to

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

Integrals

Solution:

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