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Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{0}^{\infty} \frac{x^{3}}{\left(x^{2}+1\right)^{2}} d x $$

Short Answer

Expert verified
The improper integral diverges.

Step by step solution

01

Find a comparison function and check its integral

We find a function simpler than \( \frac{x^{3}}{\left(x^{2}+1\right)^{2}} \), which in this case is \( \frac{1}{x} \). We then integrate \( \frac{1}{x} \) from 1 to infinity. The resulting improper integral \( \int_{1}^{\infty} \frac{1}{x} dx \) diverges.
02

Perform the comparison test

We can show that \( 0 \leq \frac{x^{3}}{\left(x^{2}+1\right)^{2}} \leq \frac{1}{x} \) holds for all \( x \geq 0 \). Using the comparison test, since \( \int_{1}^{\infty} \frac{1}{x} dx \) diverges, thus \( \int_{0}^{\infty} \frac{x^{3}}{\left(x^{2}+1\right)^{2}} dx \) also diverges.
03

State the conclusion

The improper integral \( \int_{0}^{\infty} \frac{x^{3}}{\left(x^{2}+1\right)^{2}} dx \) diverges. There is no finite value to evaluate.

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