Solveeit Logo
Login

Question

Mathematics Question on integral

∫01dx[ax+b(1−x)]2 \int\limits_{0} ^{1}\frac{dx}{[ax+b(1-x)]^2} is equal to

A

ab

B

ab\frac{a}{b}

C

ba\frac{b}{a}

D

1ab\frac{1}{ab}

Answer

ab\frac{a}{b}

Explanation

Solution

∫01dx[ax+b(1−x)]2=∫01[b+(a−b)x]−2dx\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 =∣b+(a−b)x−1−(a−b)∣01=\left|\frac{b+\left(a-b\right)x^{-1}}{-\left(a-b\right)}\right|_{0}^{1} =−1a−b([b+(a−b)]−1−[b]−1])=-\frac{1}{a-b} \left([b+\left(a-b\right)\right]^{-1}-\left[b\right]^{-1}]) =−1a−b[1a−1b]=-\frac{1}{a-b}\left[\frac{1}{a}-\frac{1}{b}\right] =−1a−b⋅b−aab=1ab=-\frac{1}{a-b}\cdot \frac{b-a}{ab}=\frac{1}{ab}