Given that (2x + 11 )/(2x + 1)(x + 3) ≡ A /(2x + 1) + B /(x + 3) , find the values of the constants A and B. Hence show that the integral from 0 to 2 (2x + 11)/ (2x + 1)(x + 3) dx = ln 15.

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First starting from the right hand side.

 A /(2x + 1) + B /(x + 3) = A(x+3)+B(2x+1)/(x+3)(2x+1)

Therefore the numerator = (A+2B)x+(3A+B)

Equating this numorator with the Left hand side we are presented with the two simultaneous equations A+2B=2, 3A+B=11 yielding solutions of B=-1, A=4 by elimination of A

 Hence the integral from 0 to 2  (2x + 11)/ (2x + 1)(x + 3) dx =  integral from 0 to 2 of 4/(2x+1) - 1/(x+3) dx

=[2ln(2x+1) - ln(x+3)] from 0 to 2

= [(2ln5-ln5)-(2ln1-ln3)]

=ln(5)-ln(1/3)

=ln(15)

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