Proving convergence by comparison

1. The integrand is continuous on $[0,1]$, so only $[1,+\infty)$ matters. For $x\ge1$, $x^2\ge x$ so $e^{-x^2}\le e^{-x}$, and $\int_1^\infty e^{-x}dx=e^{-1}<\infty$. The integral converges.

2. For $x\ge1$, $x^3+x\ge x^3$, hence $\frac1{\sqrt{x^3+x}}\le\frac1{x^{3/2}}$. Since $\frac32>1$, $\int_1^\infty x^{-3/2}dx$ converges. The integral converges.

3. For $x\ge2$, $\ln x<x$, so $\frac1{\ln x}>\frac1x$. Since $\int_2^\infty\frac{dx}x=+\infty$, the integral diverges.

Remark. In (1) the exact value is $\frac{\sqrt\pi}2$, but it cannot be found with an elementary antiderivative; the comparison test proves convergence without knowing the value.