Examples from Rational functions

Which of these are rational?

• $\frac{2x+1}{3x^2-4}$ is rational and proper: $\deg(2x+1)=1<2=\deg(3x^2-4)$.
• $\frac{\cos x}{x^2+1}$ is not rational: $\cos x$ is not a polynomial.
• $\frac{x^2}{x-1}$ is rational and improper: $\deg(x^2)=2>1=\deg(x-1)$.
• $\frac{\sqrt{x}}{x^2+3}$ is not rational: $\sqrt{x}$ is not a polynomial.

Polynomial plus proper remainder

Write each rational function as a polynomial plus a proper rational function.

For $\frac{4x-5}{2x+3}$, the leading-term ratio is $2$ and $2(2x+3)=4x+6$, so

Hence

For $\frac{x^3-x+1}{x^2+3x-1}$, divide. The first quotient term is $x$ and

so

The next quotient term is $-3$ and

so

Hence

Distinct linear factors

Find $\int\frac{2x^2+2x-6}{(x-1)(x+1)(x-2)}\, dx$.

The denominator is a product of distinct linear factors, so

Call this identity $(*)$. Heaviside's cover-up method finds $A$, $B$ and $C$. Multiplying $(*)$ by $x-1$ yields

Identity $(*)$ holds for $x\neq 1,-1,2$, but after multiplying through by $x-1$ both sides are defined at $x=1$, so setting $x=1$ is legitimate:

hence $A=1$.

Similarly, multiplying $(*)$ by $x+1$ and setting $x=-1$ gives

and doing the same with $x-2$ and $x=2$ gives

Hence

Repeated linear factors

Find $I=\int\frac{8x+9}{(x-2)(x+3)^2}\, dx$.

The factor $x+3$ is repeated, so

(or, equivalently, $\frac{A}{x-2}+\frac{Bx+C}{(x+3)^2}$). With the three-term form, the cover-up method from the distinct linear factors example gives $A$ by multiplying by $x-2$ and setting $x=2$:

Multiplying by $(x+3)^2$ and setting $x=-3$ gives

Cover-up does not give $B$: multiplying by $x+3$ leaves a term $\frac{C}{x+3}$, so $x=-3$ cannot be substituted. Instead, insert the known $A$ and $C$,

and set $x=0$:

Hence

This formula is valid on any interval that contains neither pole, $x=2$ nor $x=-3$; the constant need not be the same on different such intervals.

Linear numerator over a quadratic

Find $I=\int\frac{2x+1}{x^2+4x+5}\, dx$.

The quadratic $x^2+4x+5$ is irreducible ($4^2-4\cdot5=-4<0$), so write the numerator in terms of the derivative $2x+4$ of the denominator:

Thus

The first integrand is $g'/g$. Completing the square in the second, $x^2+4x+5=(x+2)^2+1$, gives

A repeated irreducible quadratic

Find $I=\int\frac{2x^2-x}{(x^2+1)^2}\, dx$.

Decompose as

Multiplying by $(x^2+1)^2$,

Comparing coefficients, $A=0$, $B=2$, $A+C=-1$ so $C=-1$ and $B+D=0$ so $D=-2$. Hence

The middle integral is $\int g'/g^2$ with $g(x)=x^2+1$, so

For the remaining integral, let $x=\tan\theta$, so $dx=\sec^2\theta\, d\theta$ (the $x=a\tan\theta$ row of the trigonometric and hyperbolic substitutions, with $a=1$). Then $\tan^2\theta+1=\sec^2\theta$ and

Using $\cos^2\theta=\frac{1+\cos 2\theta}{2}$, as in trigonometric integrals,

Altogether,

To rewrite the sine, let $\alpha=\tan^{-1}(x)$, so $\tan\alpha=x$ and $\cos\alpha>0$. From $1+\tan^2\alpha=\sec^2\alpha$ and the sign of $\cos\alpha$,

The double-angle formula $\sin(2\alpha)=2\sin\alpha\cos\alpha$ then gives

Thus