Question

1/(3x+1)^2 from 1 to inf

Answer 1

are you looking for
\[\int_1^{\infty}\frac{dx}{(3x+1)^2}\]?

Answer 2

anti - derivative is easy enough. use a u-sub
\[u=3x+1,\frac{1}{3}du=dx\] to get
\[\frac{1}{3}\int \frac{1}{u^2}du=-\frac{1}{3}\frac{1}{u}=-\frac{1}{9x+3}\]

Answer 3

yes

Answer 4

now evaluate at r and 1, you get
\[-\frac{1}{9r+3}+\frac{1}{12}\] and then take
\[\lim_{r \rightarrow \infty}-\frac{1}{9r+3}+\frac{1}{12}=0+\frac{1}{12}=\frac{1}{12}\]