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The video below goes over some examples.
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Example 1: Find $\displaystyle\int\frac6{x^2-1}\,dx$.
Solution: Since $ x^2-1= (x-1)(x+1)$, we look for a decomposition of the form $$ \displaystyle{ 6 \over (x-1)(x+1) } = \displaystyle{ { A \over x-1 } + { B \over x+1 } } $$ for some $A,\,B$. Taking the common denominator, we have $$ \displaystyle{ 6 \over (x-1)(x+1) } = \displaystyle{ { A(x+1) +B(x-1)\over (x-1)(x+1) }}. $$ This is true if the numerators are equal, i.e., $$ 6=A(x+1)+B(x-1).$$ To find out $A$, substitute $x=1$, then $6=A(2)+B(0)$, so $A=3$. Similarly, substituting $x=-1$ lets us find that $B=-3$. Finally, we have $$ \int \displaystyle{ 6 \over (x-1)(x+1) }\, dx= \int \displaystyle{ {3 \over x-1}\,dx +\int {-3 \over x+1}\,dx }=3\log{(x-1)}-3\log{(x+1)}+ C. $$ |
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Example 2: Find the partial fraction decomposition of
$$
\frac{1}{x^4-3x^3+3x^2-x}.
$$
Solution: Since $Q(x)=x\,\left(x-1\right)^3$, we have both a non-repeated factor, $x$, and a repeated factor, $(x-1)^3$, so we're looking for a decomposition of the form $$ \frac{1}{x\,\left(x-1\right)^3}=\frac{A}{x}+\frac{B_1}{x-1}+\frac{B_2}{\left(x-1\right)^2}+\frac{B_3}{\left(x-1\right)^3}. $$ Taking the common denominator, we have $$ \frac{1}{x\,\left(x-1\right)^3}=\frac{A\left(x-1\right)^3+B_1x\,(x-1)^2+B_2x\,\left(x-1\right)+B_3x} {x\left(x-1\right)^3}. $$ This is true if the numerators are equal, i.e., $$ 1=A\left(x-1\right)^3+B_1x\,(x-1)^2+B_2x\,\left(x-1\right)+B_3x.$$ To find out $B_3$, substitute $x=1$, then $1=0+B_3$, so $B_3=1$. Similarly, substituting $x=0$ let us find that $A=-1$. To find the other coefficients, we need to choose other values for $x$. For example, taking $x=2$ yields $$ 1=A+2B_1+2B_2+2B_3=-1+2B_1+2B_2+2, $$ i.e., $$ B_1+B_2=0. $$ Taking $x=-1$ yields $$ 1=-8A-4B_1+2B_2-B_3=8-4B_1+2B_2-1, $$ i.e., $$ -2B_1+B_2=-6. $$ Solving the system $$ \begin{cases} B_1+B_2=0\\ -2B_1+B_2=-6 \end{cases} $$ yields $B_1=2$ and $B_2=-2$, so that our decomposition becomes $$ \frac{1}{x\,\left(x-1\right)^3}=-\frac{1}{x}+\frac{2}{x-1}-\frac{2}{\left(x-1\right)^2}+\frac{1}{\left(x-1\right)^3}. $$ |