Under "Integral Test":
A useful result all students should work out at least once in their "calculus lives" is $ \int_1^{\infty} \frac{1}{x^p} \ dx \ , $ to see for what values of $p$ this integral converges or diverges (you may have already seen this when you covered Type I improper integrals). It will help in spotting which "p-series" $\Sigma_{n=1}^{\infty} \frac{1}{n^p} $ converge.
In your set, you'll need to use u-substitution in examining $ \int_0^{\infty} \frac{1}{(n+1)^{\gamma}} \ dx \ , $ and partial fraction decomposition for $ \int_0^{\infty} \frac{1}{(x+1) \ \cdot \ (x+2)} \ dx \ $ (or use the hint). (And having re-read your last sentences, yes, (b) and (d) converge.) $$ \\ $$
Under "Power Series" :
I believe you are correct for (d), (e) , and (f). The Ratio Test for (d) produces
$$\lim_{k \rightarrow \infty} | \ \frac{x^{2k+2}}{x^{2k}} \ \cdot \ \frac{\log (k+1)}{\log (k+2)} \ | \ = \ \lim_{k \rightarrow \infty} | \ x^2 \ \cdot \ \frac{\log (k+1)}{\log (k+2)} \ | , $$
and I don't think you need a lot of justification to declare that the limit for the ratio of logarithms is $1$ . (The ratio is equivalent to $\log_{k + 2} (k+1) $ ).
As for (e) and (f), these both hinge on dealing with $k^k$ in some manner. For (e), the ratio includes the factors
$$\frac{(k+1)! \cdot k^k}{k! \cdot (k+1)^{(k+1)}} \ , $$
which reduce to
$$(k+1) \ \cdot \ \frac{ k^k}{ (k+1)^{(k+1)}} \ = \ ( \frac{k}{ k+1})^k \ , $$
for which the limit at infinity can be found by appropriate use of l'Hopital's Rule on "indeterminate powers", or familiarity with the behavior of this function (as many members of this forum handle it).
Best Answer
You can combine the comparison test $\sum \frac{n^2}{2^n+1} < \sum \frac{n^2}{2^n}$ with the ratio test, since $$ \lim_{n\to\infty} \bigg| \frac{(n+1)^2/2^{n+1}}{n^2/2^n} \bigg| = \lim_{n\to\infty} \frac{n^2+2n+1}{2n^2} = \frac12, $$ indicating convergence. You can also use the ratio test directly on the original sum; the limit to evaluate is a bit more complicated, but still very doable.