Saturday, 18 October 2014

calculus - Evaluate inti0nftyfracdxx2+2ax+b



For $a^2



0dxx2+2ax+b




What about:




0dx(x2+2ax+b)2





My attempt is using partial fractions and completing the square, but I still failed to obtain a nice result.


Answer



Let the first integral be I and the second one be J, then by putting x=ya and y=zba2 we have
I(a,b)=0dx(x+a)2+ba2=adyy2+ba2=1ba2aba2dzz2+1=1ba2(π2arctan(aba2))

and the 2nd integral is just a first derivative of I with respect to b times 1
J(a,b)=Ib=b[1ba2(arctan(aba2)π2)]
Can you take it from here?


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