Help with improper integrals containing partial fractions Specifically,
$$\int_1^\infty \frac {24}{8x(x+1)^2} dx$$
and
$$\int_3^\infty \frac {1}{t^2 - 2t} dt$$
Both of these problems are supposed to converge, but I keep getting infinity in my answer.  For the first one, I have $$3 \lim_{b\to \infty}\left( \ln x - ln(x+1) + \frac{1}{x+1}\right)  \biggm|_{1}^b $$  and for the second I have $$\frac12 \lim_{b\to \infty} \left(\ln(t-2) - \ln(t)\right)  \biggm|_{3}^b  $$  Did I make a mistake up to that point or am I missing something with the limits?  The first answer should be $0.578$ and the second $\frac12 \ln 3$
 A: You did not make any mistakes, the thing you're missing with the limits is that $\ln a-\ln b=\ln\frac{a}{b}$.
Evaluate the limit, noting that $\ln x \to 0$ as $x \to 1$:
$$3\lim_{b\to \infty}\left( \ln x - ln(x+1) + \frac{1}{x+1}\right) \biggm|_{1}^b $$
$$= 3\lim_{b\to \infty}\left( \ln \frac{x}{x+1} + \frac{1}{x+1}\right) \biggm|_{1}^b $$
$$= 3\lim_{b\to \infty}\left( \ln \frac{b}{b+1} + \frac{1}{b+1} - \ln \frac{1}{1+1} - \frac{1}{1+1}\right) $$
$$= 3\left( 0 + 0 - \ln \frac{1}{2} - \frac{1}{2}\right) $$
$$= 3\ln 2 - \frac{3}{2}$$
Using the same method, the second limit can be evaluated.
A: Notice, 
1) $$3\lim_{b\to \infty}\left(\ln x-\ln(x+1)+\frac{1}{x+1}\right) \biggm|_{1}^{b}=3\lim_{b\to \infty}\left(\ln\left(\frac{1}{1+\frac1x}\right)+\frac{1}{x+1}\right) \biggm|_{1}^{b}$$
$$=3\lim_{b\to \infty}\left(\ln\left(\frac{1}{1+\frac 1b}\right)+\frac{1}{b+1}-\ln\left(\frac{1}{2}\right)-\frac{1}{2}\right)$$
$$=3\left(\ln\left(\frac{1}{1+0}\right)+0+\ln\left(2\right)-\frac{1}{2}\right)$$
$$=3\left(\ln\left(2\right)-\frac{1}{2}\right)=\color{red}{3\ln 2-\frac{3}{2}}$$
2) $$\frac{1}{2}\lim_{b\to \infty}\left(\ln (t-2)-\ln t\right) \biggm|_{3}^{b}=\frac{1}{2}\lim_{b\to \infty}\ln\left(\frac{t-2}{t}\right) \biggm|_{3}^{b}=\frac{1}{2}\lim_{b\to \infty}\ln\left(1-\frac{2}{t}\right) \biggm|_{3}^{b}$$
$$=\frac12\lim_{b\to \infty}\left(\ln\left(1-\frac{2}{b}\right)-\ln\left(1-\frac{2}{3}\right)\right)$$
$$=\frac12\left(\ln\left(1-0\right)-\ln\left(\frac{1}{3}\right)\right)$$
$$=\frac12\left(0+\ln(3)\right)=\color{red}{\frac{1}{2}\ln 3}$$
