# By Parts (Tabular)

## 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

Related Calculus and Beyond Homework Help News on Phys.org
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.