Interesting but tricky integration problem

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SUMMARY

The integral of the function (x^2 + 3x - 8)/(x^2 + 16) dx can be evaluated without using partial fraction expansion by first applying polynomial division. The numerator and denominator both have a degree of 2, which allows for simplification. The integral can be separated into manageable parts: S x^2/(x^2 + 16) dx, S 3x/(x^2 + 16) dx, and S -8/(x^2 + 16) dx. The integral of each part can then be computed using known integral formulas.

PREREQUISITES
  • Understanding of polynomial division
  • Familiarity with integral calculus
  • Knowledge of trigonometric integrals, specifically the formula for ∫(dx/(a^2 + x^2))
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Study polynomial long division techniques
  • Learn about integrating rational functions without partial fractions
  • Review trigonometric substitution in integrals
  • Practice solving integrals involving quadratic expressions in both the numerator and denominator
USEFUL FOR

Students and educators in calculus, particularly those tackling integration techniques and rational functions. This discussion is beneficial for anyone looking to deepen their understanding of integral calculus and polynomial manipulation.

grignard
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Homework Statement



Evaluate the integral, (x^2+3x-8/x^2+16)dx, without using partial fraction expansion. Now this has really confused me. I know that partial fraction expansion is not viable because the mumerator has a higher power overall than the denominator

The Attempt at a Solution



S x^2 /(x^2 + 16) dx + S 3x /(x^2 + 16) dx - S 8 /(x^2 + 16) dx

i have split it up like this, but i am not sure if this counts as a partial fracion expansion?

Next id use the rule, that the integral of (dx/a^2+x^2)=1/a tan(-1) x/a +c. But there are several x^2 on top, so how this does fit in.
It probably easier than it looks, any help would be much appreciaed. Thanks
 
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Use polynomial division to divide x^2+3x-8 by x^2+16 first.
 
grignard said:
I know that partial fraction expansion is not viable because the mumerator has a higher power overall than the denominator

Minor point -- the numerator does NOT have a higher power than the denominator. The degree of each is 2.
 

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