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How do I evaluate this integral? #int(2e^(2x))/(e^(2x)+16e^x+63#dx

Answer 1

Charles Arocha

#int (2e^(2x))/(e^(2x)+16e^x+63)dx = -7ln (e^x+7) + 9 ln (e^x+9) +C#

Substitute #t= e^x#, so that #dt = e^xdx#. Noting that #e^(2x) = e^x xx e^x#, we have:

#int (2e^(2x))/(e^(2x)+16e^x+63)dx = int (2t)/(t^2+16t+63) dt#

Factorize the denominator:

#t^2+16t+63 = (t+7)(t+9)#

and integrate using partial fractions decomposition:

#(2t)/(t^2+16t+63) = A/(t+7)+B/(t+9)#

#2t = A(t+9)+B(t+7)#

#2t = (A+B)t + (9A+7B)#

#{(A+B = 2),(9A+7B=0):}#

#{(A-7),(B=9):}#

#int (2t)/(t^2+16t+63) dt = -7int (dt)/(t+7) + 9 int (dt)/(t+9)#

#int (2t)/(t^2+16t+63) dt = -7ln abs (t+7) + 9 ln abs(t+9) +C#

and undoing the substitution:

#int (2e^(2x))/(e^(2x)+16e^x+63)dx = -7ln (e^x+7) + 9 ln (e^x+9) +C#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.