Evaluate the following integral ∫_0^1dx/[a_1 x+a_2 (1-x)]^2
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total answers (1)
Mathematics
I=∫_0^1dx/[a_1 x+a_2 (1-x)]^2
need an explanation for this answer? contact us directly to get an explanation for this answer=∫_0^1[(a_1-a_2 )x+a_2 ]^(-2) dx=∫_0^1〖[(a_1-a_2 )x+a_2 ]^(-2)/((a_1-a_2 ) ) (a_1-a_2 )dx〗
I=1/((a_1-a_2 ) ) [-((a_1-a_2 )x+a_2 )^(-1) ]_0^1=1/((a_1-a_2 ) ) [1/a_2 -1/(a_1-a_2+a_2 )]
=1/((a_1-a_2 ) ) [(a_1-a_2)/(a_1 a_2 )]=1/(a_1 a_2 )