A) \[\frac{a}{b}\]
B) \[\frac{b}{a}\]
C) \[a\,b\]
D) \[\frac{1}{a\,b}\]
Correct Answer: D
Solution :
Let \[I=\int_{0}^{1}{\frac{dx}{{{[(a-b)x+b]}^{2}}}}\] Put \[t=(a-b)x+b\Rightarrow dt=(a-b)dx\] As \[x=1\Rightarrow t=a\]and \[x=0\Rightarrow t=b\], then \[I=\frac{1}{a-b}\int_{b}^{a}{\frac{1}{{{t}^{2}}}}dt=\frac{1}{(a-b)}\left[ -\frac{1}{t} \right]_{b}^{a}=\frac{1}{(a-b)}\left( \frac{a-b}{ab} \right)=\frac{1}{ab}\].You need to login to perform this action.
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