Solveeit Logo
Login

Question

Mathematics Question on Integrals of Some Particular Functions

3a∫01(ax−1a−1)2  dx3a{{\int_{0}^{1}{\left( \frac{ax-1}{a-1} \right)}}^{2}}\,\,dx is equal to

A

a−1+(a−1)−2a-1+{{(a-1)}^{-2}}

B

a+a−2a+{{a}^{-2}}

C

a−a−2a-{{a}^{-2}}

D

a2+1a2{{a}^{2}}+\frac{1}{{{a}^{2}}}

Answer

a−1+(a−1)−2a-1+{{(a-1)}^{-2}}

Explanation

Solution

3a∫01(ax−1a−1)2 dx=3a(a−1)2[(ax−1)33×1a]013a\int_{0}^{1}{{{\left( \frac{ax-1}{a-1} \right)}^{2}}\,dx=\frac{3a}{{{(a-1)}^{2}}}}\left[ \frac{{{(ax-1)}^{3}}}{3}\times \frac{1}{a} \right]_{0}^{1}
=1(a−1)2[(a−1)3+1]=\frac{1}{{{(a-1)}^{2}}}[{{(a-1)}^{3}}+1]
=(a−1)+(a−1)−2=(a-1)+{{(a-1)}^{-2}}