Integrate rational functions

Rational logarithmic forms and partial fractions

1. Compare the numerator and denominator degrees; divide first if needed.
2. Factor the denominator and include every repeated-factor power.
3. Solve for the partial-fraction coefficients, then integrate each term.
4. Keep the denominator restrictions and differentiate the final answer to check it.

For $a>0$, the source gives $\int1/(x^2-a^2)\,dx=\frac1{2a}\ln[(x-a)/(x+a)]+C$ on $x>a$, and its composite version $\int f^{\prime}(x)/([f(x)]^2-a^2)\,dx=\frac1{2a}\ln[(f(x)-a)/(f(x)+a)]+C$ on $f(x)>a$. It also gives $\int1/(a^2-x^2)\,dx=\frac1{2a}\ln[(a+x)/(a-x)]+C$ on $|x|<a$, and the composite version $\int f^{\prime}(x)/(a^2-[f(x)]^2)\,dx=\frac1{2a}\ln[(a+f(x))/(a-f(x))]+C$ on $|f(x)|<a$. The basic formulas are marked MF27. The Teacher highlights using moduli inside logarithms when a sign restriction is not specified. With moduli these formulas hold on any connected interval avoiding the denominator zeros; moduli do not remove poles.

The source recalls three partial-fraction patterns (with distinct factors and nonzero denominators):

• $\frac{px+q}{(ax+b)(cx+d)}=\frac A{ax+b}+\frac B{cx+d}$.
• $\frac{px^2+qx+r}{(ax+b)(cx+d)^2}=\frac A{ax+b}+\frac B{cx+d}+\frac C{(cx+d)^2}$: retain every power of a repeated linear factor.
• $\frac{px^2+qx+r}{(ax+b)(x^2+c^2)}=\frac A{ax+b}+\frac{Bx+C}{x^2+c^2}$: an irreducible quadratic needs a linear numerator.

The Teacher emphasises starting with a proper rational fraction and recognising factor structure; divide first if it is improper. The source labels these templates MF27.

To derive the first logarithmic formula, write $1/(x^2-a^2)=P/(x-a)+Q/(x+a)$. Clearing denominators gives $1=P(x+a)+Q(x-a)$. Substituting $x=a$ gives $P=1/(2a)$ and $x=-a$ gives $Q=-1/(2a)$.

Thus integration gives $[\ln|x-a|-\ln|x+a|]/(2a)+C=\frac1{2a}\ln\left|\frac{x-a}{x+a}\right|+C$.

The displayed logarithm law is $\ln a-\ln b=\ln(a/b)$ for positive arguments; applying it to absolute values preserves the needed signs. Self-Practice 6 asks for the companion proof.

If a numerator is not directly proportional to the denominator derivative, split it into that derivative plus a remainder. Example 9 uses $6x+16=3(2x+4)+4$. Example 10 preserves two methods and the Teacher note explaining why their different-looking logarithms agree.

Check the logarithm and its interval

A student claims $\int x/(x^2-4)\,dx=\ln(x^2-4)+C$. Differentiate the proposed answer to test it, then give a corrected antiderivative. State the intervals on which your real-valued answer applies and explain what happens between the two excluded inputs.

After your attempt: compare the reasoning

The proposed derivative is $2x/(x^2-4)$, twice the required integrand. The corrected form is

$\frac12\ln|x^2-4|+C.$

It differentiates to $x/(x^2-4)$ on each of $(-\infty,-2)$, $(-2,2)$ and $(2,\infty)$. On the middle interval, $x^2-4<0$, so the real logarithm can be written $\ln(4-x^2)$. Its derivative still gives the required sign after the chain rule. The integrand excludes $x=\pm2$; the constants may be chosen independently on its separate intervals.