Solve Integral [xln(x^2+9)] Using Tabular Method

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SUMMARY

The integral of [xln(x^2+9)] with respect to x can be solved using the tabular method, which involves integration by parts. The user initially set u as ln(x^2+9) and dv as x, leading to a partial solution of ((1/2)x^2)ln(x^2+9) - (1/3)[integral of](x^4/(x^2+9))dx. The discussion highlights a common error in the integration process, specifically the misapplication of the tabular method when the degree of the numerator exceeds that of the denominator. The correct approach involves adjusting the u and dv selections to avoid complications with trigonometric functions.

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  • Knowledge of logarithmic functions and their properties
  • Basic skills in polynomial long division
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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!
 
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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.
 
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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