Edited to write the improper limits as limits of proper integrals, in response to Arturo's comment.
a) The integrand $\log x$ has a singular point at $x=0$. The improper integral of the second king $I=\int_{0}^{1}\log x\;\mathrm{d}x$ is, by definition, the limit
$\lim_{\varepsilon \rightarrow 0^{+}}\int_{\varepsilon }^{1}\log x\;\mathrm{% d}x.$
The integral $\int \log x\;\mathrm{d}x$ is usually integrated by parts
$\begin{eqnarray*} I &=&\lim_{\varepsilon \rightarrow 0^{+}}\int_{\varepsilon }^{1}\log x\;% \mathrm{d}x=\lim_{\varepsilon \rightarrow 0^{+}}\int_{\varepsilon }^{1}1\cdot \log x\;\mathrm{d}x \\ &=&\lim_{\varepsilon \rightarrow 0^{+}}\left[ x\log x\right] _{\varepsilon }^{1}-\lim_{\varepsilon \rightarrow 0^{+}}\int_{\varepsilon }^{1}x\cdot \frac{1}{x}\;\mathrm{d}x \\ &=&1\cdot \log 1-\lim_{\varepsilon \rightarrow 0^{+}}\varepsilon \log \varepsilon -1 =-\lim_{\varepsilon \rightarrow 0^{+}}\varepsilon \log \varepsilon -1=-1, \end{eqnarray*}$
where $\lim_{\varepsilon \rightarrow 0^{+}}\varepsilon \log \varepsilon $ was evaluated by l'Hôpital's rule:
$\lim_{\varepsilon \rightarrow 0^{+}}\varepsilon \log \varepsilon =\lim_{\varepsilon \rightarrow 0^{+}}\frac{\log \varepsilon }{\frac{1}{% \varepsilon }}=\lim_{\varepsilon \rightarrow 0^{+}}\frac{\frac{1}{% \varepsilon }}{-\frac{1}{\varepsilon ^{2}}}=\lim_{\varepsilon \rightarrow 0^{+}}-\varepsilon =0.$
Added:
b) The integral $\int_{a}^{+\infty }\frac{1}{x^{p}}\;\mathrm{d}x$ is divergent for $a>0,p\leq 1$, as can be seen by evaluating it. We apply the limit test to $f(x)=\frac{\log x}{x% }$ and $g(x)=\frac{1}{x}$:
$\frac{f(x)}{g(x)}=\frac{\log x}{x}\cdot x=\log x\rightarrow \infty,\qquad\text{ as } x\rightarrow \infty .$
Both $f(x)$ and $g(x)$ are nonnegative functions in $[2,+\infty \lbrack $. Since $\int_{2}^{\infty }g(x)\;\mathrm{d}x=\int_{2}^{\infty }\frac{1}{x}\;\mathrm{d}x$ is divergent, so is $\int_{2}^{\infty }f(x)\;\mathrm{d}x=\int_{2}^{\infty }\frac{\log x}{x}\;% \mathrm{d}x.$
c) The improper integral $I=\int_{0}^{\infty }\frac{1}{1+x^{2}}\;\mathrm{d}x$ is of the first kind, because the integrand has no singularities. By definition of an integral of such a kind, it is the limit
$\lim_{b\rightarrow +\infty }\int_{0}^{b}\frac{1}{1+x^{2}}\;\mathrm{d}x.$
Since $\int \frac{1}{1+x^{2}}\;\mathrm{d}x=\arctan x$, we have: $\begin{eqnarray*} I &=&\lim_{b\rightarrow +\infty }\int_{0}^{b}\frac{1}{1+x^{2}}\;\mathrm{d}x =\lim_{b\rightarrow +\infty }\left[ \arctan x\right] _{0}^{b} \\ &=&\lim_{b\rightarrow +\infty }\arctan b-\arctan 0 =\frac{\pi }{2}-0=\frac{\pi }{2}. \end{eqnarray*}$