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$$\int_{2} ^{4}\dfrac{dx}{x(\ln x)^2}.$$

Here is what I did:

$$u=\ln x, du=\dfrac{dx}{x}$$

$$\int_{2} ^{4}u^{-2}du$$

$$(-1)u^{-1} |_{2}^{4}$$

$$-\dfrac{1}{\ln x}|_{2}^{4}$$

$$-\dfrac{1}{\ln 4} + \dfrac{1}{\ln 2}$$

However the answer in the back of my textbook says that the answer is $\dfrac{1}{\ln 4}$. I have went over my work a couple of times and I cannot see what I did wrong. Could someone please explain what's wrong here? Thank you.

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By a common denominator, this is the same as

$$\frac{-\ln{2} + \ln{4}}{\ln{4} \ln{2}} = \frac{-\ln{\frac{1}{2}}}{\ln{4} \ln{2}} = \frac{\ln{2}}{\ln{4} \ln{2}} = \frac{1}{\ln{4}}$$

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  • $\begingroup$ I am not sure how you got your numerator in your second step. If I applied the quotient rule for logs It would go straight to the numerator of your third step. I wasn't aware that I could simplify my answer. $\endgroup$ – Kot Sep 27 '13 at 1:08
  • $\begingroup$ @StevenN $\ln{a} - \ln{b} = \ln{a} + \ln{1/b} = \ln{a/b}$ $\endgroup$ – user61527 Sep 27 '13 at 1:10
  • $\begingroup$ I still can't figure out why $-ln\dfrac{1}{2}$ is $ln2$. Could you please explain? $\endgroup$ – Kot Sep 27 '13 at 1:15
  • $\begingroup$ @StevenN It is a property of logarithms that $-\ln{a} = \ln{1/a}$. $\endgroup$ – user61527 Sep 27 '13 at 1:17
  • $\begingroup$ Interesting... I do not recall learning that property. Thanks for the clear explanation! $\endgroup$ – Kot Sep 27 '13 at 1:22
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That's right! $$-\frac{1}{ln(2²)}+\frac{1}{ln(2)}=-\frac{1}{2ln(2)}+\frac{1}{ln(2)}=1/2\frac{1}{ln(2)}=\frac{1}{ln(2²)}$$ You will always find answers like this.

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