Asked by Ashwin
Evaluate Integral x^2+4 / x^4+x^2+16 dx
Answers
Answered by
Steve
gotta get creative with this one. Note that
x^4+x^2+16 = x^4-8x^2+16 - 9x^2 = (x^2-4)^2 - (3x)^2
= (x^2-4)^2 [(3x/(x^2-4)^2)^2 + 1]
so, if you let u = 3x/(x^2-4)
du = -3(x^2+4)/(x^2-4)^2 and you have
integral -1/3 du/(1+u^2)
take it from there!
x^4+x^2+16 = x^4-8x^2+16 - 9x^2 = (x^2-4)^2 - (3x)^2
= (x^2-4)^2 [(3x/(x^2-4)^2)^2 + 1]
so, if you let u = 3x/(x^2-4)
du = -3(x^2+4)/(x^2-4)^2 and you have
integral -1/3 du/(1+u^2)
take it from there!
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.