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By Parts (Tabular)

  • Thread starter phisci
  • Start date
  • #1
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Homework Statement


Solve the integral of [xln(x^2+9)] wrt x using the tabular method.


Homework Equations


By parts using the tabular method.


The Attempt at a Solution



u:
1. ln(x^2+9)
2. 2x/(x^2+9)

dv:
1. x
2. (1/2)x^2
3. (1/6)x^3

The answer for now is : ((1/2)x^2)ln(x^2+9) -(1/3)[integral of](x^4/(x^2+9))dx

dividing x^4/(x^2+9) gives (x^2)-9+(81/x^2+9)

however integrating 1/(x^2+9) would give a trig function whereas the answer given contains to trig function at all. Where have I gone wrong in my method? Thanks in advance!
 

Answers and Replies

  • #2
It's easier to integrate x^4 than it is to integrate 1/(x^2+9). Set your u and dv values differently and apply tabular integration once again.
 
  • #3
10
0
Hi thanks for the reply. I've tried your method but it gives back the same problem. As in the exponent of x in the numerator is greater than that of the exponent of x in the denominator. My initial approach to this was to carry out long division however I got a [81/(x^2+9)] which yields a trig function when integrated. However the ans to this question does not contain any trig function. Where have I gone wrong? Thanks.
 

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