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MATH
[1251]

Methods from Rational functions

All methods from Rational functions.

Back to Rational functions

Partial fraction terms

Given a proper rational function, each irreducible factor of the denominator contributes terms as follows.

  • Linear factor ax+bax+b: Aax+b\frac{A}{ax+b}.
  • Repeated linear factor (ax+b)k(ax+b)^k: A1ax+b+A2(ax+b)2+⋯+Ak(ax+b)k\frac{A_1}{ax+b}+\frac{A_2}{(ax+b)^2}+\cdots+\frac{A_k}{(ax+b)^k}.
  • Irreducible quadratic factor ax2+bx+cax^2+bx+c with a≠0a\ne0: Ax+Bax2+bx+c\frac{Ax+B}{ax^2+bx+c}.
  • Repeated irreducible quadratic factor (ax2+bx+c)k(ax^2+bx+c)^k: A1x+B1ax2+bx+c+A2x+B2(ax2+bx+c)2+⋯+Akx+Bk(ax2+bx+c)k\frac{A_1x+B_1}{ax^2+bx+c}+\frac{A_2x+B_2}{(ax^2+bx+c)^2}+\cdots+\frac{A_kx+B_k}{(ax^2+bx+c)^k}.

Here each factor has a general leading coefficient aa, whereas the decomposition theorem used the monic quadratic x2+bx+cx^2+bx+c, irreducible when b2−4c<0b^2-4c<0.

Integrate a rational function

To integrate a rational function on an interval containing no zero of its denominator:

  1. If the function is improper, use polynomial division to write it as a polynomial plus a proper rational function and integrate the polynomial.
  2. Decompose the proper part into partial fractions.
  3. Integrate each linear-factor term using u=ax+bu=ax+b: its first power gives Aaln⁡∣ax+b∣\frac{A}{a}\ln|ax+b|, while a power k>1k>1 gives A(ax+b)1−ka(1−k)\frac{A(ax+b)^{1-k}}{a(1-k)}.
  4. For q(x)=ax2+bx+cq(x)=ax^2+bx+c and a term ∫Bx+Cq(x)k dx\int\frac{Bx+C}{q(x)^k}\,dx, split the numerator into a multiple of q′(x)=2ax+bq'(x)=2ax+b and a constant remainder:
Bx+Cq(x)k=B2aq′(x)q(x)k+Dq(x)k,D=C−Bb2a.\frac{Bx+C}{q(x)^k} =\frac{B}{2a}\frac{q'(x)}{q(x)^k}+\frac{D}{q(x)^k}, \qquad D=C-\frac{Bb}{2a}.

Then integrate the derivative term directly and complete the square in the constant remainder.

The derivative term is immediate:

∫g′(x)g(x) dx=ln⁡∣g(x)∣+K,\int\frac{g'(x)}{g(x)}\, dx=\ln|g(x)|+K,

and, for k>1k>1,

∫g′(x)g(x)k dx=g(x)1−k1−k+K.\int\frac{g'(x)}{g(x)^k}\,dx =\frac{g(x)^{1-k}}{1-k}+K.

For the constant term, complete the square. When k=1k=1, this produces the standard inverse-tangent integral

∫1h2+x2 dx=1htan⁡−1(xh)+K,h>0.\int\frac{1}{h^2+x^2}\, dx=\frac{1}{h}\tan^{-1}\left(\frac{x}{h}\right)+K, \qquad h>0.

For k>1k>1, reduce the remaining power or use a trigonometric substitution, as in the repeated-quadratic example below.

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