Integration Partial Fractions

The technique of partial fractions actually does not introduce any new calculus techniques at all, but is just a nice application of algebra to the integration of rational functions, (recall a rational function is the quotient of two polynomials). This collection of strategies all aims to replace a single rational function with a sum of simpler rational functions.
It’s easiest to motivate with an example: the integral $\int \frac{1}{x}+\frac{1}{x-2}dx$ is easy to compute, as we can break into the sum of two integrals: the first of which we recognize and the second of which turns into the same with a quick $u$-substituion $$\int\frac{1}{x}dx+\int\frac{1}{x-2}dx=\ln|x|+\ln|x-2|+C$$

We can disguise this easy integral as a more difficult one, if we add together the fractions first before integrating: $$\frac{1}{x}+\frac{1}{x-2}=\frac{x-2}{x(x-2)}+\frac{x}{x(x-2)}=\frac{2x-2}{x^2-2x}$$

If we were given the integral $\int \frac{2x-2}{x^2-2x}dx$ to compute, we might be initially at a loss, as $u$-substitution, trigonometric substitution and integration by parts all fail to make meaningful progress. But if we could somehow see how to “un-add” these two fractions, and convert the integrand into a simpler state, we would realize this integral is no trouble at alL! Partial fractions is the technique that allows us to do exactly this.

The Main Strategy

How can we do this methodically? Say for sake of specificity that we are looking at a rational function $p(x)/q(x)$ where $p$ is linear (either a constant, or $ax+b$) and the denominator $q$ is quadratic.

Important Cases

Numerator is Higher Degree than the Denominator

If this is the case, we can always first do polynomial long division to replace the rational function $p/q$ with the sum of some polynomial (which is easy to integrate!) and a new rational function, where the numerator (the remainder of the long division) has a degree smaller than the denominator.

Example: $$\frac{x^3-2x+1}{x^2-4}=x+\frac{2x+1}{x^2-4}$$

Denominator Contains all Distinct Linear Factors

If the denominator factors as a product of distinct linear factors, we will always succeed at representing this rational function as a sum of very simple pieces: each term will be a constant divided by one of the linear factors. In the quadratic case, this was the foundational example we saw above. But the same holds in higher degree: for instance we can write

$$\frac{x^2-x+1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$$

Denominator Contains a Repeated Factor

If the denominator contains a repeated factor (something like $(x-3)^2$), the partial fractions decomposition requires that we include terms for all powers of this denominator, up to and including the power that appears in the original. That is, if we have a rational function like

$$\frac{3x-2}{x^3(1+x)}$$

the partial fractions decomposition is

$$\frac{3x-2}{x^3(1+x)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{1+x}$$

Denominator Contains an Irreducible Quadratic

Every real polynomial factors into a combination of linear factors and quadratic factors such as $x^2+1$ that don’t factor any further. Because these cant be reduced to a product of linear polynomials, tehy’re called irreducible quadratics, and when they show up in the denominator of a rational function, the partial fractions decomposition has a linear term in the numerator. For example, $$\frac{x}{x(x-1)(x^2+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{Cx+D}{x^2+1}$$

When you find the correct constants and then go to integrate, you’ll be confronted with an integral of the form $$\int\frac{Cx+D}{x^2+1}dx$$

(Cases where the denominator contains repeated copies of the same irreducible quadratic are possible, but could not possibly appear unless the denominator had degree 4 or higher….thus we will not see them often in class, and need not worry about them here).

Example Problems

Distinct Factors

$$\int\frac{3x^2+2x+1}{x^3-x}dx$$ Factor denominator into $x(x+1)(x-1)$, set up partial fractions system.

Repeated Factors

$$\int \frac{x(3-5x)}{(3x-1)(x-1)^2}dx$$ Here the denominator contains a repeated term, so the system includes terms with both $(x-1)$ and $(x-1)^2$.

Irreducible Quadratics

$$\int \frac{x+5}{x(x^2+1)}dx$$ Here we have an irreducible quadratic, so we need a linear term in the numerator of one of our components. This will lead to us doing a u-sub and remembering the integral $\arctan(x)=\int \frac{dx}{1+x^2}$$

The partial fractions decomposition is $$\frac{A}{x}+\frac{Bx+C}{x^2+1}$$ Adding fractions gives a new numerator of $$A(x^2+1)+(Bx+C)x$$ which must be equal to the original numerator of $x+5$. We get a system of equations by substituting in values for $x$: if $x=0$ this becomes $A=5$. If $x=1$ this becomes $2A+B+C=6$ and if $x=-1$ this becomes $2A-B+C=4$. Solving this system we see $C=-5$ and $B=1$. Thus $$\int\frac{x+5}{x^2+1}dx=\int\frac{5}{x}+\frac{x-5}{x^2+1}dx$$ To integrate this, we break up the final term into a sum of two fractions, and do all three integrals separately: $$5\int\frac{1}{x}dx+\int\frac{x}{x^2+1}dx-5\int\frac{1}{x^2+1}dx$$ The first is $\ln|x|$, the last is $\arctan(x)$, and the middle one is accomplished via the $u$-sub $u=x^2+1$.

Irreducible quadratics in Disguise

What happens if we see a denominator like $3x^2+4$? For example, just in the integral $\int\frac{1}{3x^2+4}$, what can we do? First - recall how to derive the following with $u$-sub: $$\int\frac{1}{u^2+a^2}du=\frac{1}{a}\arctan\frac{u}{a}+C$$ Then, we can do the $u$-sub with $u=3x$ to get $$\int\frac{1}{3x^2+4}dx=\frac{1}{3}\int\frac{1}{u^2+4}du=\frac{1}{3}\left(\frac{1}{2}\arctan\frac{x}{2}\right)+C$$

What about an irreducible quadratic like $2x^2-4x+10$? Here we can complete the square and do a $u$-sub, just like we saw while doing trigonometric substitutions. $$2x^2-4x+10=2(x^2-2x+5)=2((x-1)^2-1+5)=2((x-1)^2+4)=2(x-1)^2+8$$ So, $$\int\frac{1}{2x^2-4x+10}dx=\left(\int \frac{1}{\sqrt{8}}\arctan\frac{x-1}{\sqrt{8}}\right)+C$$ After doing a $u$-sub for $u=x-1$ and then doing the procedure above.

Substitution Leads to Partial Fractions

Rationalizing via a u-substitution $$\int\frac{\sqrt{2x+1}{x}}dx$$ Do sub $u=\sqrt{2x+1}$ so $x=\frac{1}{2}(u^2-1)$ and $dx=udu$. Then this becomes $$2\int \frac{u^2}{u^2-1}du=2\int 1+\frac{1}{u^2-1}du$$ And we can do standard partial fractions.

$$\int\frac{dx}{x\sqrt{x-1}}$$ Do the substitution $u=\sqrt{x-1}$.

$$\int\frac{\sqrt{1+\sqrt{x}}{x}dx$$ Do the substitution $u=\sqrt{1+\sqrt{x}}$ This leads to $u^2-1=\sqrt{x}$ so $x=(u^2-1)^2$ and $du=\frac{1}{2\sqrt{1+\sqrt{x}}\frac{1}{2\sqrt{x}}dx$$ so $$4u\sqrt{x}du=dx$, or $4u(u^2-1)du=dx$. Putting all this together, we see the integral becomes $$\int\frac{u}{(u^2-1)^2}4u(u^2-1)du=4\int\frac{u^2}{u^2-1}du$$ this is the same integral we saw in the last problem!

$$\int\sec\theta d\theta$$ Multiply by $\cos\theta /\cos\theta$, convert denominator into $1-\sin^2\theta$, do a $u$-sub, do partial fractions.

$$\int \frac{e^x}{(e^x-2)(e^{2x}+1)dx$$