Methods from Rational functions
All methods from Rational functions.
[METHOD]Given a proper rational function, each irreducible factor of the denominator contributes terms as follows.
- Linear factor ax+b: ax+bA.
- Repeated linear factor (ax+b)k: ax+bA1+(ax+b)2A2+⋯+(ax+b)kAk.
- Irreducible quadratic factor ax2+bx+c with a=0: ax2+bx+cAx+B.
- Repeated irreducible quadratic factor (ax2+bx+c)k: ax2+bx+cA1x+B1+(ax2+bx+c)2A2x+B2+⋯+(ax2+bx+c)kAkx+Bk.
Here each factor has a general leading coefficient a, whereas the decomposition theorem used the monic quadratic x2+bx+c, irreducible when b2−4c<0.
[METHOD]To integrate a rational function on an interval containing no zero of its
denominator:
- If the function is improper, use polynomial division to write it as a polynomial plus a proper rational function and integrate the polynomial.
- Decompose the proper part into partial fractions.
- Integrate each linear-factor term using u=ax+b: its first power gives aAln∣ax+b∣, while a power k>1 gives a(1−k)A(ax+b)1−k.
- For q(x)=ax2+bx+c and a term
∫q(x)kBx+Cdx, split the numerator into a multiple of
q′(x)=2ax+b and a constant remainder:
q(x)kBx+C=2aBq(x)kq′(x)+q(x)kD,D=C−2aBb.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,and, for k>1,
∫g(x)kg′(x)dx=1−kg(x)1−k+K.For the constant term, complete the square. When k=1, this produces the
standard inverse-tangent integral
∫h2+x21dx=h1tan−1(hx)+K,h>0.For k>1, reduce the remaining power or use a trigonometric substitution, as
in the repeated-quadratic example below.