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## Linear FactorsLet $f(x)$ be the proper rational function
$f(x)=\frac{P(x)}{Q(x)}$. To use the method of partial
fractions to integrate $f(x)$, we first factor $Q(x)$. In
the case when
The video below goes over some examples. ## More Examples
----------------------------------------------------------------------------------Example 2: Find the partial fraction decomposition of $$
\frac{1}{x^4-3x^3+3x^2-x}. $$ DO: The issue here is factoring a
quartic, which can be quite difficult. Hint:
$Q(x)=x\,\left(x-1\right)^3$. Work on this before
proceeding.Solution 2: 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 $x=0$ we 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 $$ \frac{1}{x\,\left(x-1\right)^3}=-\frac{1}{x}+\frac{2}{x-1}-\frac{2}{\left(x-1\right)^2}+\frac{1}{\left(x-1\right)^3}. $$ ----------------------------------------------------------------------------------Example 3: DO: Use your answer above to find $\displaystyle\int\frac{1}{x(x-1)^3}\,dx$. Solution 3:
$\displaystyle\int\frac{1}{x\,\left(x-1\right)^3}\,dx=-\int\frac{1}{x}\,dx+\int\frac{2}{x-1}\,dx-\int\frac{2}{\left(x-1\right)^2}\,dx+\int\frac{1}{\left(x-1\right)^3}\,dx$$\displaystyle=-\ln|x|+2\ln|x-1|+\frac{2}{(x-1)}-\frac{1}{2(x-1)^2}+C$, using $u$-substitution on the last two terms. Differentiate to see if this answer is right.DO: |