Asked by byu
                (e^(3x)-2e^(2x)+(5e^x))/(e^x+1) how do I do this what is the techniques??
            
            
        Answers
                    Answered by
            Steve
            
    I'd start with a long division. The integrand then becomes
8 - 3e^x + e^(2x) - 8/(e^x+1)
8 - 8/(e^x+1) = e^x/(e^x+1)
if you let
u = e^x+1
du = e^x dx
and the integrand is now
-3e^x + e^(2x) - 8du/u
making the integral
-3e^x + 1/2 e^(2x) - 8log(e^x+1) + C
    
8 - 3e^x + e^(2x) - 8/(e^x+1)
8 - 8/(e^x+1) = e^x/(e^x+1)
if you let
u = e^x+1
du = e^x dx
and the integrand is now
-3e^x + e^(2x) - 8du/u
making the integral
-3e^x + 1/2 e^(2x) - 8log(e^x+1) + C
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