# Thread:

1. ## Clarification on Definite Integrals

Hello, and thanks in advance for your help.

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.

2. integrate ln(x)/x^2 - Wolfram|Alpha

click show steps on the right to see the integration by parts

3. Originally Posted by NBrunk
Hello, and thanks in advance for your help.

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 $\ln(x)=z$.

4. 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!

5. 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 $-\frac{1}{2}\ln 2+1-\frac{1}{2}=\frac{1-\ln 2}{2}.$

6. ## Correction and Restatement of Question

Pending Correction

7. 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
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