Limit as $t \rightarrow\infty$ of $(\ln(x+5)/26)-(\ln(t^2+1)/52)+(5\tan^{-1}(t))/26)-(\ln(5)/26)$

Question

I had this integral,

$$\int_0^{\infty} \frac{dx}{(x+5) \,(x^2 + 1)}$$ I simplified using partial fractions and eventually got to $$\lim_{t \rightarrow\infty} \frac{\ln(t+5)}{26}-\frac{\ln(t^2+1)}{52} +\frac{\ 5arctan(t)}{26} - \frac{\ln5}{26}$$

At this step, after plugging in infinity, I get $$ \infty - \infty \, + \frac{\pi}2- \frac{\ln 5}{26}$$

Clearly, in the form infinity - infinity, you cannot apply L'Hopital's Rule. Is there anyway to simplify this to $$\frac{\infty}{\infty}$$

so that the Rule can be applied?

Answer 1

Hint

$$\frac{\log(t+5)}{26}-\frac{\ln(t^2+1)}{52}=\frac 1{52}\left(\log((t+5)^2)-\log(t^2+1) \right)=\frac 1{52}\log\left(\frac{(t+5)^2}{t^2+1}\right)$$ could help