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$$\int\frac{x^5+x-1}{x^3 +1} dx$$

Have tried everything ... polynomial long division, partial fractions, trig substitution etc...

Not for an assignment, so if a complete solution could be provided that'd be much appreaciated

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Hint:

$$\frac{x^5+x-1}{x^3 +1} = \frac{(x^3+x^2-1)(x^2-x+1)}{(x+1)(x^2-x+1)}$$

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  • $\begingroup$ Would the first step not be to use polynomial long division? $\endgroup$ – Guest May 27 '14 at 0:10
  • $\begingroup$ @Guest Do you want the first step to be polynomial division? $\endgroup$ – David H May 27 '14 at 0:13
  • $\begingroup$ Well, generally partial fractions is utilised when degree of the numerator is less than the degree of the denominator $\endgroup$ – Guest May 27 '14 at 0:28
  • $\begingroup$ @Guest Sure, just like how you were taught to use long division to divide two numbers and find the remainder when the given some fraction like $\frac{1210}{55}$. How would you go about simplifying that fraction? $\endgroup$ – David H May 27 '14 at 0:36
  • $\begingroup$ Using the division algorithm .... $\endgroup$ – Guest May 27 '14 at 0:39
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First divide the denominator into the numerator to get

$$\frac{x^5+x-1}{x^3+1}=x^2 -\frac{x^2-x+1}{x^3+1}$$

Now we factor the denominator, $$x^3+1=(x+1)(x^2-x+1)$$

Note that we have cancellation so we get $$\int x^2-\frac{1}{x+1} dx=\frac{x^3}{3}-\ln |x+1|+C$$

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  • $\begingroup$ Thanks a lot ... can't believe I missed such a simple detail on cancelling out like terms $\endgroup$ – Guest May 27 '14 at 0:12
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    $\begingroup$ I didnt see it either, you just need to follow the method. $\endgroup$ – Rene Schipperus May 27 '14 at 0:30

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