# 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 $B$.

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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 $B$.

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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}

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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.

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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 $B$).

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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. $$A=-B=\frac38.$$

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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 $$x-4=0 \rightarrow x=4 \Rightarrow \qquad A=\frac{3}{x+4}\Big |_{x=4}=\frac{3}{8}$$ $$x+4=0 \rightarrow x=-4 \Rightarrow \qquad B=\frac{3}{x-4}\Big |_{x=-4}=\frac{3}{-8}$$

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