Asked by Jane
                Use integration by parts to find the integral. Round the answer to two decimal places if necessary. (x+4) ln x dx between 3 and 0.
I tried this problem two different ways and got two different answers 1.63 and 3.89. which one is correct? Please.
            
        I tried this problem two different ways and got two different answers 1.63 and 3.89. which one is correct? Please.
Answers
                    Answered by
            Steve
            
    u = ln x
du = 1/x dx
dv = (x+4) dx
v = x^2/2 + 4x
Int[(x+4)ln x dx] = (x^2/2 + 4x)lnx - Int(x/2 + 4)dx
= (x^2/2 + 4x)lnx - (x^2/4 + 4x)
Evaluating over the interval, we get
[9/2 + 12)ln3 - (9/4 + 12)] - [0]
= 33/2 ln3 - 57/4
= 3.8771
    
du = 1/x dx
dv = (x+4) dx
v = x^2/2 + 4x
Int[(x+4)ln x dx] = (x^2/2 + 4x)lnx - Int(x/2 + 4)dx
= (x^2/2 + 4x)lnx - (x^2/4 + 4x)
Evaluating over the interval, we get
[9/2 + 12)ln3 - (9/4 + 12)] - [0]
= 33/2 ln3 - 57/4
= 3.8771
                    Answered by
            Jane
            
    same thing I got, but not even close to any of the answer choices ... 2.69, 16.69, -2.15, and 1.51. This one has me stumped!!
    
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