Determine if the integral of x^3/(x^5 + 2)dx from [1,infinity) converges or diverges.

Hint: Use the Limit Comparison Test with a suitable function.

If anyone can help me take the integral of this problem, I would be very greatful.  Thank you anyone who took the time to look over my question.

We can write the original expression as x^3/(x^3(x^2+2/x^3)=

1/(x^2+2/x^3)=1/(x^2(1+2x^-5))=x^-2(1+2x^-5)^-1=

x^-2(1-2x^-5+4x^-10-8x^-15-16x^-20+...)=

(x^-2)-(2x^-7)+(4x^-12)-... . When this is integrated, we get -x^-1+(x^-6)/3-(4x^-11)/11+(x^-16)/2-... Applying the limits we get 1-1/3+4/11-1/2+16/21-16/13+...+(-2)^n/(5n+1)+... .

If we take terms in pairs apart from the first we have:

1+(4/11-1/3)+(16/21-1/2)+(64/31-16/13)+...=1+1/33+11/42+336/403+... [The actual formula for these pair differences is (2^(2(m-1)))(10m-9)/((10m+1)(5m-2)), where n=2m, the strictly positive even values of n. When, for example, m=1, this expression evaluates to 1/33.] These fractions are getting progressively larger, because of the powers of 2 in the numerator, so the series is diverging. That means the integral is divergent towards infinity.

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