# Clarification on Definite Integrals

• January 21st 2010, 06:02 PM
NBrunk
Clarification on Definite Integrals

The problem I'm facing is to integrate by parts the following function:

lnx/x^2 with the interval of integration being from 1 to 2.

I don't think I need a full solution, it's just that after I do the integration by parts method, I'm left with no integrals, and this makes me concerned. Is the answer to this problem simply my simplified 1/3x^3lnx - 1/9x^3 or is it this simplification with the values 2 inputted for x subtracted from it with 1 inputted for x (like shown below)?

[1/3*2^3*ln2 - 1/9*2^3] - [1/3*1^3*ln1 - 1/9*1^3]

Thanks again for the assistance...I think it's just a matter of not understanding how the interval of integration given at the start of the problem relates to my solution.
• January 21st 2010, 06:04 PM
emathinstruction
integrate ln(x)/x^2 - Wolfram|Alpha

click show steps on the right to see the integration by parts
• January 21st 2010, 06:05 PM
Drexel28
Quote:

Originally Posted by NBrunk

The problem I'm facing is to integrate by parts the following function:

lnx/x^2 with the interval of integration being from 1 to 2.

I don't think I need a full solution, it's just that after I do the integration by parts method, I'm left with no integrals, and this makes me concerned. Is the answer to this problem simply my simplified 1/3x^3lnx - 1/9x^3 or is it this simplification with the values 2 inputted for x subtracted from it with 1 inputted for x (like shown below)?

[1/3*2^3*ln2 - 1/9*2^3] - [1/3*1^3*ln1 - 1/9*1^3]

Thanks again for the assistance...I think it's just a matter of not understanding how the interval of integration given at the start of the problem relates to my solution.

Try letting $\ln(x)=z$.
• January 21st 2010, 06:24 PM
NBrunk
Thanks for the link, it showed how wrong my solution was...took dv = simply x^2 rather than x^-2 and it threw me off.

However, the link shows the solution to the indefinite integral, and I'm still not sure how to reflect the definite interval of integration of 1 to 2 in my answer...although I'm much closer!
• January 21st 2010, 06:49 PM
Krizalid
Okay listen (actually read); we want to compute $\int_1^2\frac{\ln x}{x^2}\,dx.$

You already spotted that this needs to be integrated by parts, now the trick here, it's to express $\ln x$ or $\frac1{x^2}$ as a function being differentiated, but you'll probably say "I don't get your point." It's simple, let's think: do we have a function whose derivative gives me $\ln x$ ? Yes, it does exist, but, do you know it that fast so that you can tell me what it is? - No, so let's think on $\frac1{x^2},$ we can immediately see that $\left(-\frac1x\right)'=\frac1{x^2},$ so we have $\int_{1}^{2}{\frac{\ln x}{x^{2}}\,dx}=\int_{1}^{2}{\left( -\frac{1}{x} \right)'\ln (x)\,dx}.$ This will be equal to the following: multiply $-\frac{1}{x}$ and $\ln x$ evaluated for $1\le x\le2$ and the remaining integral will be $-\int_1^2\frac1{x}(\ln x)'\,dx,$ so our original integral becomes $\left. -\frac{1}{x}\ln x \right|_{1}^{2}+\int_{1}^{2}{\frac{1}{x}\left( \ln x \right)'\,dx}=-\frac{1}{2}\ln 2+\int_{1}^{2}{\frac{dx}{x^{2}}}=-\frac{1}{2}\ln 2+\left( \left. -\frac{1}{x} \right|_{1}^{2} \right).$

So our integral equals $-\frac{1}{2}\ln 2+1-\frac{1}{2}=\frac{1-\ln 2}{2}.$
• January 21st 2010, 06:50 PM
NBrunk
Correction and Restatement of Question
Pending Correction
• January 21st 2010, 08:11 PM
Pulock2009
1) let ln x=t
2) int(e^t.t)dt
3) by parts integration
4) use ILATE rule
5) ans:e^t(t-1)
6) ans in terms of x:x(ln x-1)
7) put the limits