How to analyze convergence or divergence of the integral $\int_1^{\infty}(t^2+\ln^2t)^{-1}dt$

Question

Analyze convergence or divergence of the integral $\displaystyle\int_1^{\infty}(t^2+\ln^2t)^{-1}dt$

since $\displaystyle\int f(y)^{-1}dy=yf(y)^{-1}-F(f(y)^{-1})+C$

$\displaystyle\int(t^2+\ln^2t)^{-1}dt=t(t^2+\ln^2t)^{-1}-F((t^2+\ln^2t)^{-1})+C $

to find $\int(t^2+\ln^2t)^{-1}dt$ $\;\;\;\;t=e^k$ $\quad dt=e^kdk$

$\int (e^{3k}+k^2e^k)dk$

should I continue in this way?

Answer 1

It's much easier to just proceed with a comparison test directly: The term $\ln^2 t$ is very small when compared to $t^2$, so it's almost irrelevant. More precisely, we have

$$t^2 < t^2 + \ln^2 t \implies \frac{1}{t^2 + \ln^2 t} < \frac{1}{t^2}$$

for all $t > 1$. Now using the fact that

$$\int_1^{\infty} \frac{1}{t^2} dt = - \frac 1 t \Big|_{1}^{\infty} = 1$$

is convergent, we're done.