FORM GR1268 Question 3

Question:
\(\displaystyle \int_{e^{-3}}^{e^{-2}}\frac{1}{x\log x}dx=\)

(A) \(1\)     (B) \(\displaystyle\frac{2}{3}\)     (C) \(\displaystyle\frac{3}{2}\)     (D) \(\displaystyle\log\left(\frac{2}{3}\right)\)     (E) \(\displaystyle\log\left(\frac{3}{2}\right)\)

Answer:
(D)

Answer Key:
\[\begin{align}
\int_{e^{-3}}^{e^{-2}}\frac{1}{x\log x}dx
&= \left.\log\left(\log x\right)\right|_{e^{-3}}^{e^{-2}}\\
&=\log\left(\log(e^{-2})\right) - \log\left(\log(e^{-3})\right)\\
&=\log(-2)-\log(-3)\\
&=\log\left(\frac{-2}{-3}\right)\\
&=\log\left(\frac{2}{3}\right)
\end{align}\]

I implicitly used the substitution rule.

Write
\[\int_{e^{-3}}^{e^{-2}}\frac{1}{x\log x}dx
=\int_{e^{-3}}^{e^{-2}}\frac{1}{x}\cdot\frac{1}{\log x}dx
=\int_{e^{-3}}^{e^{-2}}(\log x)'\cdot\frac{1}{\log x}dx.\]

Let \(u=\log x\). Then \(du=\frac{1}{x}dx\).
The limits of integration in terms of \(u\) are \(\log(e^{-3})=-3\) and \(\log(e^{-2})=-2\).

Thus,
\[\begin{align}
\int_{e^{-3}}^{e^{-2}}\frac{1}{x\log x}dx
&=\int_{e^{-3}}^{e^{-2}}\frac{1}{\log x}\frac{1}{x}dx\\
&=\int_{-3}^{-2}\frac{1}{u}du\\
&=\left.\log u\right|_{-3}^{-2}\\
&=\log(-2)-\log(-3)\\
&=\log\left(\frac{-2}{-3}\right)\\
&=\log\left(\frac{2}{3}\right).
\end{align}\]