3.4 Partial fractions  (Page 2/5)

Since $\text{degree}(x^2+3x+1)\ge \text{degree}(x^2-4),$ we must perform long division of polynomials. This results in

Next, we perform partial fraction decomposition on $\frac{3x+5}{x^2-4}=\frac{3x+5}{(x+2)(x-2)}.$ We have

Thus,

Solving for $A$ and $B$ using either method, we obtain $A=11\text{/}4$ and $B=1\text{/}4.$

Rewriting the original integral, we have

Evaluating the integral produces

Let’s begin by letting $u=\text{sin}x.$ Consequently, $du=\text{cos}xdx.$ After making these substitutions, we have

Applying partial fraction decomposition to $1\text{/}u(u-1)$ gives $\frac{1}{u(u-1)}=-\frac{1}{u}+\frac{1}{u-1}.$

Thus,

We have $\text{degree}\left(x-2\right)<\text{degree}\left({\left(2x-1\right)}^2\left(x-1\right)\right),$ so we can proceed with the decomposition. Since ${(2x-1)}^2$ is a repeated linear factor, include $\frac{A}{2x-1}+\frac{B}{{(2x-1)}^2}$ in the decomposition. Thus,

After getting a common denominator and equating the numerators, we have

We then use the method of equating coefficients to find the values of $A,$ $B,$ and $C.$

Equating coefficients yields $2A+4C=0,$ $\mathrm{-3}A+B-4C=1,$ and $A-B+C=\mathrm{-2}.$ Solving this system yields $A=2,$ $B=3,$ and $C=\mathrm{-1}.$

Alternatively, we can use the method of strategic substitution. In this case, substituting $x=1$ and $x=1\text{/}2$ into [link] easily produces the values $B=3$ and $C=\mathrm{-1}.$ At this point, it may seem that we have run out of good choices for $x,$ however, since we already have values for $B$ and $C,$ we can substitute in these values and choose any value for $x$ not previously used. The value $x=0$ is a good option. In this case, we obtain the equation $\mathrm{-2}=A\left(\mathrm{-1}\right)\left(\mathrm{-1}\right)+3\left(\mathrm{-1}\right)+(\mathrm{-1}){(\mathrm{-1})}^2$ or, equivalently, $A=2.$

Now that we have the values for $A,$ $B,$ and $C,$ we rewrite the original integral and evaluate it: