Examples

Partial fractions with repeated factor

Find the partial fraction decomposition:

$$\frac{3x-9}{x^3+3x^2-4}$$

Solution

(1) Check that denominator degree is lower. $✓$

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(2) Factor denominator:

Rational Roots Theorem: check for roots at $\pm 1$ and $\pm 2$ and $\pm 4$.

Discover that $x=+1$ is a root. Therefore divide by $x-1$:

$$\frac{x^3+3x^2-4}{x-1}\gg \gg x^2+4x+4$$

Factor again:

$$x^2+4x+4\gg \gg {(x+2)}^2$$

Final factored form:

$$x^3+3x^2-4=(x-1){(x+2)}^2$$

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(3) Write the generic PFD:

$$\frac{3x-9}{(x-1){(x+2)}^2}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{{(x+2)}^2}$$

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(4) Solve for $A$, $B$, and $C$:

Multiply across by the common denominator:

$$3x-9=A{(x+2)}^2+B(x-1)(x+2)+C(x-1)$$

For $A$, set $x=1$, obtain:

$$\begin{array}{cccc}& & 3\cdot 1-9& =A{(1+2)}^2+B\cdot 0+C\cdot 0\\ \gg \gg & & -6& =9A\\ \gg \gg & & A& =-2/3\end{array}$$

For $C$, set $x=-2$, obtain:

$$\begin{array}{cccc}& & 3\cdot (-2)-9& =A\cdot 0+B\cdot 0+C\cdot (-3)\\ \gg \gg & & -15& =-3C\\ \gg \gg & & C& =5\end{array}$$

For $B$, insert prior results and solve.

Plug in $A$ and $C$:

$$3x-9=-\frac{2}{3}{(x+2)}^2+B(x-1)(x+2)+5(x-1)$$

Now plug in another convenient $x$, say $x=3$:

$$\begin{array}{cc}0& =-\frac{2}{3}\cdot 5^2+B\cdot 2\cdot 5+5\cdot 2\\ \frac{50}{3}-10& =10B\gg \gg B=\frac{2}{3}\end{array}$$

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(4) Plug in $A$, $B$, $C$ for the final answer:

$$\frac{3x-9}{x^3+3x^2-4}=\frac{-2/3}{x-1}+\frac{2/3}{x+2}+\frac{5}{{(x+2)}^2}$$

Partial fractions - repeated quadratic, linear tops

Compute the integral:

$$\int \frac{x^3+1}{{(x^2+4)}^2}dx$$

Solution

(1) Partial fraction decomposition:

• Numerator degree is lower than denominator.
• Factor denominator completely. (No real roots.)

Write generic PFD:

$$\frac{x^3+1}{{(x^2+4)}^2}=\frac{Ax+B}{x^2+4}+\frac{Cx+D}{{(x^2+4)}^2}$$

• Notice: repeated factor: use incrementing powers up to 2.
• Notice: linear over quadratic.

Common denominators and solve:

$$\begin{array}{c}{x}^3+1=(Ax+B)(x^2+4)+Cx+D\\ \\ \gg \gg x^3+1=A{x}^3+B{x}^2+(4A+C)x+4B+D\\ \\ \gg \gg A=1,B=0\\ \\ \gg \gg C=-4,D=1\end{array}$$

Therefore:

$$\frac{x^3+1}{{(x^2+4)}^2}=\frac{x}{x^2+4}+\frac{-4x+1}{{(x^2+4)}^2}$$

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(2) Integrate:

Integrate the first term using substitution $u=x^2+4$:

$$\begin{array}{c}\int \frac{x}{x^2+4}dx{\gg \gg }^{u=x^2+4}\frac{1}{2}\int \frac{du}{u}\\ \\ \gg \gg \frac{1}{2}\ln |u|+C\gg \gg \frac{1}{2}\ln |x^2+4|+C\end{array}$$

Break up the second term:

$$\frac{-4x+1}{{(x^2+4)}^2}\gg \gg \frac{-4x}{{(x^2+4)}^2}+\frac{1}{{(x^2+4)}^2}$$

Integrate the first term of RHS:

$$\begin{array}{c}\int \frac{-4x}{{(x^2+4)}^2}dx\gg \gg -2\int \frac{du}{u^2}\\ \\ \gg \gg \frac{2}{u}+C\gg \gg \frac{2}{x^2+4}+C\end{array}$$

Integrate the second term of RHS using $x=2\tan \theta$:

$$\begin{array}{c}\int \frac{dx}{(x^2+4{)}^2}\gg \gg \int \frac{2{\sec }^2\theta d\theta }{16{\sec }^4\theta }\\ \\ \gg \gg \frac{1}{8}\int {\cos }^2\theta d\theta \\ \\ \gg \gg \frac{1}{8}\int \frac{1}{2}(1+\cos (2\theta ))d\theta \\ \\ \gg \gg \frac{1}{16}\theta +\frac{1}{32}\sin (2\theta )+C\\ \\ \gg \gg \frac{1}{16}{\tan }^{-1}\left(\frac{x}{2}\right)+\frac{1}{32}2\sin \theta \cos \theta +C\\ \\ \gg \gg \frac{1}{16}{\tan }^{-1}\left(\frac{x}{2}\right)+\frac{1}{32}2\frac{x}{\sqrt{x^2+2^2}}\frac{2}{\sqrt{x^2+2^2}}+C\\ \\ \gg \gg \frac{1}{16}{\tan }^{-1}\left(\frac{x}{2}\right)+\frac{1}{8}\frac{x}{x^2+2^2}+C\end{array}$$