Since the numerator has higher degree than the denominator, start by dividing: $\displaystyle \frac{x^4}{x^2+ 1}= x^2- 1- \frac{1}{x^2+ 1}$. The first two should be easy and $\displaystyle \int \frac{1}{x^2+ 1} dx= arctan(x)+ C$

Apr 22nd 2010, 10:05 AM

AlphaRock

Quote:

Originally Posted by HallsofIvy

Since the numerator has higher degree than the denominator, start by dividing: $\displaystyle \frac{x^4}{x^2+ 1}= x^2- 1- \frac{1}{x^2+ 1}$. The first two should be easy and $\displaystyle \int \frac{1}{x^2+ 1} dx= arctan(x)+ C$

Thanks. Now if the numerator has the same degree as the denominator, would we just leave and do the integration by partial fractions?

Apr 22nd 2010, 02:45 PM

craig

Quote:

Originally Posted by AlphaRock

Thanks. Now if the numerator has the same degree as the denominator, would we just leave and do the integration by partial fractions?

If the numerator and denominator have the same degree, it's technically still an improper fraction so cancelling it will often make in easier to integrate.