Int by parts - tabular method

[markraz]

Hi how do you do this problem by tabular method?


\(\displaystyle \int x^2 ln(x) \)

I can do it by standard methods but I can't seem to figure it out with tabular

what I a have so far:

	derive	integrate
+	ln(x)	x^2
-	1/x	(x^3)/3
+	-1/x^2	(x^4)/12

put it together

[ln(x) ((x^3)/3)] - [(1/x)((x^4)/12)]


but the answer is:
[ln(x) ((x^3)/3)] - [(x^3)/9)]+C

any thoughts ?

 

[Ishuda]

Hi how do you do this problem by tabular method?


\(\displaystyle \int x^2 ln(x) \)

I can do it by standard methods but I can't seem to figure it out with tabular

what I a have so far:

	derive	integrate
+	ln(x)	x^2
-	1/x	(x^3)/3
+	-1/x^2	(x^4)/12

put it together

[ln(x) ((x^3)/3)] - [(1/x)((x^4)/12)]


but the answer is:
[ln(x) ((x^3)/3)] - [(x^3)/9)]+C

any thoughts ?

You either keep going until you get a zero or you are left with the final 'u dv'. So you are left with the final with that final integral of (-1/x^2)(x^4/12) or -x^2/12 which integrates to -x^3/36. The 1/36 plus the 1/12 is 4/36 or 1/9.

 

[markraz]

You either keep going until you get a zero or you are left with the final 'u dv'. So you are left with the final with that final integral of (-1/x^2)(x^4/12) or -x^2/12 which integrates to -x^3/36. The 1/36 plus the 1/12 is 4/36 or 1/9.

thanks so I should keep going?