Question

Use the comparison Theorem to determine whether the integral

\(\int\limits_0^\infty {\frac{{{x^3}}}{{{x^5} + 2}}dx} \)is convergent or divergent.

Short answer

Hence proved the given expression\(\int\limits_1^\infty {\frac{{{x^3}}}{{{x^5} + 2}}dx} \)is convergent

Step-by-step solution

Step 1: Applied the interval

\({x^5} + 2 > {x^5}\)on the interval

\(\begin{aligned}{l}{x^5} + 2 > {x^5}\\\frac{1}{{{x^5} + 2}} < \frac{1}{{{x^5}}}\\\frac{{{x^3}}}{{{x^5} + 2}} < \frac{{{x^3}}}{{{x^5}}}\\ = \frac{1}{{{x^2}}}\end{aligned}\)

Step 2: Use the comparison theorem to solve the expression

Ifon the intervalthenconverges ifis convergent

\(\int\limits_1^\infty {\frac{{{x^3}}}{{{x^5} + 2}}dx < \int\limits_1^\infty {\frac{1}{{{x^2}}}dx} } \)

Step 3: Final Proof

\(\int\limits_1^\infty {\frac{1}{{{x^p}}}} dx\)is convergent for

Hence\(\int\limits_1^\infty {\frac{1}{{{x^2}}}} dx\)is convergent.

\(\int\limits_1^\infty {\frac{{{x^3}}}{{{x^5} + 2}}dx} = \) convergent