Show that $$\int_1^2\cfrac {dx}{x\ln^2 x} \qquad \qquad \qquad (a)$$ Converges or diverges ,
I can think of using a substitution, $u=\ln x$ , $du=\cfrac 1xdx$
\begin{align} & = \int_1^2\cfrac {dx}{x\ln^2 x}\\ & = \int_1^2\cfrac {du}{u^2}\\ & = \lim \limits_{c \to 1^+}\int_c^2\cfrac {du}{u^2} \\ & = \lim \limits_{c \to 1^+}[-1u^{-1}]|_c^2 \\ & = -\lim \limits_{c \to 1^+}[\frac 12-\frac1c ] \\ & = \frac 12 \\ \end{align} I am right? or what i've done wrong?
We are asked to evaluate $\int_{1}^{2} \frac{1}{x(\ln (x))^2}dx$.
As you suggested, change of variables is a good approach to this problem. Going by your suggestion, let $u(x) = \ln x$. It follows, $du = \frac{1}{x}dx$. However, we need to make changes to the limits of integrations too! That is the key. So $u(1) = \ln(1)=0$ and $u(2)=\ln(2)$. Therefore,
$$ \int_{1}^{2} \frac{1}{x(\ln (x))^2}dx = \lim \limits_{c \rightarrow 0^+} \int_{c}^{\ln(2)} \frac{1}{xu^2}xdu = \lim \limits_{c \rightarrow 0^+} \int_{c}^{\ln(2)} \frac{1}{u^2}du $$
$$ = \lim \limits_{c \rightarrow 0^+} \int_c^{\ln(2)} u^{-2} du = \lim \limits_{c \rightarrow 0^+} (-\frac{1}{3u}) \big|_c^{\ln(2)} = (-\frac{1}{3\ln(2)})+ \lim \limits_{c \rightarrow 0^+} \frac{1}{3c}$$
The first term is a finite number and the second term goes to $+\infty$.
Hence, the integral diverges to $ \infty $.
