# Help with integration using partial fractions

I'm not sure how to get the values for $A$ and $B$ for the expression $$\frac{3}{x^2 - 16}.$$ I've split the expression into $$\frac{A}{x - 4} + \frac{B}{x + 4}.$$ I don't know what to do afterwards to get the values for $A$ and $B$.

-

Multiply both sides of the equation $$\frac{3}{x^2 - 16} = \frac{A}{x - 4} + \frac{B}{x + 4}$$ by $(x - 4)(x + 4)$ to clear denominators. Now you have the polynomial equation $$3 = A(x + 4) + B(x - 4),$$ which must hold for all values of $x$. By evaluating at $x = 4$, you can find $A$, and by evaluating at $x = -4$, you can find $B$.

-
It only works with linear factors of multiplicity $1$, though. – Sammy Black Dec 18 '13 at 22:36
@SammyBlack: True, but if dealing say with denominator say $(x-1)^2(x-2)$ we can use the trick to get $2$ of the coefficients, and then the third one is easy. – André Nicolas Dec 18 '13 at 23:21

$$\frac{3}{x^2-16} = \frac{A}{x-4}+\frac{B}{x+4}$$ $$3 = A(x+4) + B(x-4)$$ $$0=(A+B)x + (4A - 4B - 3)$$ \begin{cases}A+B = 0 \\ 4A-4B-3 = 0\end{cases}

-

Set the two expressions equal to each other, multiply by $x^2-16$, and gather like terms with a $0$ on one side of the equation. The coefficients of the polynomial on the other side should be equal to $0,$ which gives you a solvable linear system.

-

Get a common denominator, add them, and equate the numerator to $3$. The numerator will have a constant term and an $x$ term, something like $Ux+V$.You want this to be equal to $3$, or in other words, $0x+3$. Set $U=0$ and $V=3$. This gives a system of 2 equations which you can solve to get $A$ and $B$. (Note: $U$ and $V$ will be linear combinations of $A$ and $B$).

-

Since $$\frac{A}{x-4}+\frac{B}{x+4}=\frac{(A+B)x+4(A-B)}{x^2-16},$$ we have $$\frac{3}{x^2-16}=\frac{A}{x-4}+\frac{B}{x+4} \iff A+B=0,\quad 4(A-B)=3,$$ i.e. $$A=-B=\frac38.$$

-

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{{3 \over x^{2} - 16} = {A_{+} \over x - 4} + {A_{-} \over x + 4}}$ $$A_{\pm} = \lim_{x \to \pm 4}\bracks{\pars{x \mp 4}\,{3 \over x^{2} - 16}} =\lim_{x \to \pm 4}\pars{3 \over 2x} = \pm\,{3 \over 8} \quad\imp\quad \left\lbrace \begin{array}{rcr} A_{+} & = & {3 \over 8} \\ A_{-} & = & -\,{3 \over 8} \end{array}\right.$$ Then $$\color{#0000ff}{\large {3 \over x^{2} - 16}} = \color{#0000ff}{\large{3/8 \over x - 4} - {3/8 \over x + 4}}$$

-

There is a trick for finding such numbers. $$\frac{3}{x^2-16}=\frac{3}{(x-4)(x+4)}=\frac{A}{x-4}+\frac{B}{x+4}$$ then $$x-4=0 \rightarrow x=4 \Rightarrow \qquad A=\frac{3}{x+4}\Big |_{x=4}=\frac{3}{8}$$ $$x+4=0 \rightarrow x=-4 \Rightarrow \qquad B=\frac{3}{x-4}\Big |_{x=-4}=\frac{3}{-8}$$

-