Which of the following integrals is unbounded? A. \[\int\limits_0^{\frac{\pi }{4}} {\tan {\text{x dx}}} \] B. \[\int\limits_0^\infty {\frac{1}{{{{\text{x}}^2} + 1}}{\text{dx}}} \] C. \[\int\limits_0^\infty {{\text{x}}{{\text{e}}^{ – {\text{x}}}}{\text{ dx}}} \] D. \[\int\limits_0^1 {\frac{1}{{1 – {\text{x}}}}{\text{dx}}} \]

” option2=”\[\int\limits_0^\infty {\frac{1}{{{{\text{x}}^2} + 1}}{\text{dx}}} \]” option3=”\[\int\limits_0^\infty {{\text{x}}{{\text{e}}^{ – {\text{x}}}}{\text{ dx}}} \]” option4=”\[\int\limits_0^1 {\frac{1}{{1 – {\text{x}}}}{\text{dx}}} \]” correct=”option2″]

The correct answer is $\boxed{\text{(B)}}$.

The integral $\int_0^\infty \frac{1}{{x^2} + 1} \, dx$ is unbounded because the integrand function $\frac{1}{{x^2} + 1}$ does not approach zero as $x$ approaches infinity. In fact, the function $\frac{1}{{x^2} + 1}$ oscillates between positive and negative values infinitely many times as $x$ approaches infinity, so the integral diverges.

The integral $\int_0^{\frac{\pi}{4}} \tan x \, dx$ is bounded because the integrand function $\tan x$ is bounded on the interval $[0, \frac{\pi}{4}]$. The integral $\int_0^\infty x e^{-x} \, dx$ is also bounded because the integrand function $xe^{-x}$ is decreasing and approaches zero as $x$ approaches infinity. The integral $\int_0^1 \frac{1}{1-x} \, dx$ is also bounded because the integrand function $\frac{1}{1-x}$ is decreasing and approaches a positive value as $x$ approaches zero.