Convergence or Divergence integral test $$\sum_{n=1}^\infty\frac1{(\sqrt n + n )\ln (\sqrt n+1)}$$ 
So im guessing you go $\frac{1}{f(x)}$ and $\frac{1}{g(x)}$
and see if $\frac{1}{f(x)}$ and $\frac{1}{g(x)}$ are both convergent by finding the antiderivative of both.
Is that the right way to do it? It doesn't seem to be the right way for as I go around in circles doing it this way.
Thanks for the help

Note that $\sqrt{n} < n$. Hence, $n+\sqrt{n} < 2n \implies \dfrac1{n+\sqrt{n}} > \dfrac1{2n}$.
Similarly, $\sqrt{n}+1 < n$ for $n>2$. Hence, $\log(\sqrt{n}+1) < \log n \implies \dfrac1{\log(\sqrt{n}+1)} > \dfrac1{\log n}$.
This gives us that
$$\dfrac1{n+\sqrt{n}} \dfrac1{\log(1+\sqrt{n})} > \dfrac1{2n \log n}$$
Now use integral test to conclude that $\displaystyle \sum_{n=1}^{\infty} \dfrac1{2n \log n}$ diverges and hence $$\displaystyle \sum_{n=1}^{\infty} \dfrac1{n+\sqrt{n}} \dfrac1{\log(1+\sqrt{n})}$$ diverges.
