7.5: Partial Fractions

Since \(\deg(3x+2) \lt \deg(x^3-x^2-2x)\), we begin by factoring the denominator of \(\frac{3x+2}{x^3-x^2-2x}\).

We can see that \(x^3-x^2-2x=x(x-2)(x+1)\). Thus, there are constants \(A\), \(B\), and \(C\) satisfying Equation \ref{eq:7.4.1} such that\[\dfrac{3x+2}{x(x-2)(x+1)}=\dfrac{A}{x}+\dfrac{B}{x-2}+\dfrac{C}{x+1}.\nonumber\]We must now find these constants. To do so, we begin by multiplying both sides of this equation by the least common denominator: \(x(x - 2)(x + 1)\). Thus,\[3x+2=A(x-2)(x+1)+Bx(x+1)+Cx(x-2).\label{Ex2Numerator}\]There are two different strategies for finding the coefficients \(A\), \(B\), and \(C\). We refer to these as the method of equating coefficients and the method of strategic substitution.

Since \(\deg(x^2+3x+1) \geq \deg(x^2-4)\), we must perform long division of polynomials. This results in\[\dfrac{x^2+3x+1}{x^2-4}=1+\dfrac{3x+5}{x^2-4}.\nonumber\]Next, we perform partial fraction decomposition on \(\frac{3x+5}{x^2-4}=\frac{3x+5}{(x+2)(x-2)}\). We have\[\dfrac{3x+5}{(x-2)(x+2)}=\dfrac{A}{x-2}+\dfrac{B}{x+2}.\nonumber\]Thus,\[3x+5=A(x+2)+B(x-2).\nonumber\]Solving for \(A\) and \(B\) using either method, we obtain \(A= \frac{11}{4}\) and \(B= \frac{1}{4}\).

Rewriting the original integral, we have\[\int \dfrac{x^2+3x+1}{x^2-4}\,dx=\int \left(1+\dfrac{11}{4} \cdot \dfrac{1}{x-2}+\dfrac{1}{4} \cdot \dfrac{1}{x+2}\right)\,dx.\nonumber\]Evaluating the integral produces\[\int \dfrac{x^2+3x+1}{x^2-4}\,dx=x+\dfrac{11}{4}\ln |x-2|+\dfrac{1}{4}\ln |x+2|+C.\nonumber\]

Let's begin by letting \(u=\sin x\). Consequently, \(du=\cos x\,dx\). After making these substitutions, we have\[\int \dfrac{\cos x}{\sin^2x-\sin x}\,dx=\int \dfrac{du}{u^2-u}=\int \dfrac{du}{u(u-1)}.\nonumber\]Applying partial fraction decomposition to \(\frac{1}{u(u-1)}\) gives \(\frac{1}{u(u-1)}=-\frac{1}{u}+\frac{1}{u-1}\).

Thus,\[\int \dfrac{\cos x}{\sin^2x-\sin x}\,dx=-\ln |u|+\ln |u-1|+C=-\ln |\sin x|+\ln |\sin x-1|+C.\nonumber\]

We have \(\deg(x-2) \lt \deg((2x-1)^2(x-1))\), so we can proceed with the decomposition. Since \((2x-1)^2\) is a repeated linear factor, include\[\dfrac{A}{2x-1}+\dfrac{B}{(2x-1)^2}\nonumber\]in the decomposition in Equation \ref{eq:7.4.2}. Thus,\[\dfrac{x-2}{(2x-1)^2(x-1)}=\dfrac{A}{2x-1}+\dfrac{B}{(2x-1)^2}+\dfrac{C}{x-1}.\nonumber\]After multiplying both sides by the least common denominator, we have\[x-2=A(2x-1)(x-1)+B(x-1)+C(2x-1)^2. \label{Ex5Numerator}\]We then use the method of equating coefficients to find the values of \(A, B\), and \(C\).\[x-2=(2A+4C)x^2+(-3A+B-4C)x+(A-B+C).\nonumber\]Equating coefficients yields \(2A+4C=0\), \(-3A+B-4C=1\), and \(A-B+C=-2\). Solving this system yields \(A=2\), \(B=3\), and \(C=-1\).

Alternatively, we can use the method of strategic substitution. In this case, substituting \(x=1\) and \(x= \frac{1}{2}\) into Equation \(\ref{Ex5Numerator}\) easily produces the values \(B=3\) and \(C=-1\). 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 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 \(-2=A(-1)(-1)+3(-1)+(-1)(-1)^2\) or, equivalently, \(A=2\).

Now that we have the values for \(A\), \(B\), and \(C\), we rewrite the original integral and evaluate it:\[\begin{array}{rcl} \displaystyle \int \dfrac{x-2}{(2x-1)^2(x-1)}\,dx & = & \displaystyle \int \left(\dfrac{2}{2x-1}+\dfrac{3}{(2x-1)^2}-\dfrac{1}{x-1}\right)\,dx \\[16pt] & = & \ln |2x-1|-\dfrac{3}{2(2x-1)}-\ln |x-1|+C. \\[16pt] \end{array}\nonumber\]

Since \(\deg(2x-3) \lt \deg(x^3+x)\), factor the denominator and proceed with partial fraction decomposition. In this case, \(x^3+x=x(x^2+1)\) contains the irreducible quadratic factor \(x^2+1\). Therefore, we include \(\frac{Ax+B}{x^2+1}\) as part of the decomposition, along with \(\frac{C}{x}\) for the linear term \(x\). Thus, the decomposition has the form\[\dfrac{2x-3}{x(x^2+1)}=\dfrac{Ax+B}{x^2+1}+\dfrac{C}{x}.\nonumber\]After multiplying both sides by the least common denominator, we obtain the equation\[2x-3=(Ax+B)x+C(x^2+1).\nonumber\]Solving for \(A\), \(B\), and \(C\), we get \(A=3\), \(B=2\), and \(C=-3\).

Thus,\[\dfrac{2x-3}{x^3+x}=\dfrac{3x+2}{x^2+1}-\dfrac{3}{x}.\nonumber\]Substituting back into the integral, we obtain\[\begin{array}{rcl} \displaystyle \int \dfrac{2x-3}{x^3+x}\,dx & = & \displaystyle \int \left(\dfrac{3x+2}{x^2+1}-\dfrac{3}{x}\right)\,dx \\[16pt] & = & 3 \displaystyle \int \dfrac{x}{x^2+1}\,dx + 2 \displaystyle \int \dfrac{1}{x^2+1}\,dx-3 \displaystyle \int \dfrac{1}{x}\,dx \\[16pt] & = & \dfrac{3}{2}\ln ∣x^2+1∣+2 \tan^{-1}x-3\ln |x|+C. \\[16pt] \end{array}\nonumber\]Note: We may rewrite \(\ln ∣x^2+1∣=\ln (x^2+1)\), if we wish to do so, since \(x^2+1 \gt 0\); however, this is not required.

We can start by factoring \(x^3-8=(x-2)(x^2+2x+4)\). We see that the quadratic factor \(x^2+2x+4\) is irreducible since \(2^2-4(1)(4) = -12 \lt 0\). Using the decomposition described in the problem-solving strategy, we get\[\dfrac{1}{(x-2)(x^2+2x+4)}=\dfrac{A}{x-2}+\dfrac{Bx+C}{x^2+2x+4}.\nonumber\]Multiplying both sides by the LCD, we arrive at\[1=A(x^2+2x+4)+(Bx+C)(x-2).\nonumber\]Applying either method to find the constants, we get \(A=\frac{1}{12}\), \(B=-\frac{1}{12}\), and \(C=-\frac{1}{3}\).

Rewriting \(\displaystyle \int \frac{dx}{x^3-8}\), we have\[\int \dfrac{\,dx}{x^3-8}=\dfrac{1}{12}\int \dfrac{1}{x-2}\,dx-\dfrac{1}{12}\int \dfrac{x+4}{x^2+2x+4}\,dx.\nonumber\]We can see that\[\int \dfrac{1}{x-2}\,dx=\ln |x-2|+C,\nonumber\]but\[\int \dfrac{x+4}{x^2+2x+4}\,dx\nonumber\]requires a bit more effort. Let's begin by completing the square on \(x^2+2x+4\) to obtain\[x^2+2x+4=(x+1)^2+3.\nonumber\]By letting \(u=x+1\) and consequently \(du=\,dx\), we see that\[\begin{array}{rcl} \displaystyle \int \dfrac{x+4}{x^2+2x+4}\,dx & = & \displaystyle \int \dfrac{x+4}{(x+1)^2+3}\,dx \\[16pt] & = & \displaystyle \int \dfrac{u+3}{u^2+3}\,du \\[16pt] & = & \displaystyle \int \dfrac{u}{u^2+3}\,du + \displaystyle \int \dfrac{3}{u^2+3}\, du \\[16pt] & = & \dfrac{1}{2}\ln ∣u^2+3∣+\dfrac{3}{\sqrt{3}}\tan^{-1}\left(\dfrac{u}{\sqrt{3}}\right)+C \\[16pt] & = & \dfrac{1}{2}\ln ∣x^2+2x+4∣+\sqrt{3}\tan^{-1}\left(\dfrac{x+1}{\sqrt{3}}\right)+C \\[16pt] \end{array}\nonumber\]Substituting back into the original integral and simplifying gives\[\int \dfrac{dx}{x^3-8}=\dfrac{1}{12}\ln |x-2|-\dfrac{1}{24}\ln |x^2+2x+4|-\dfrac{\sqrt{3}}{12}\tan^{-1}\left(\dfrac{x+1}{\sqrt{3}}\right)+C.\nonumber\]Here again, we can drop the absolute value if we wish to do so, since \(x^2+2x+4 \gt 0\) for all \(x\).

Let's begin by sketching the region to be revolved (see Figure \(\PageIndex{1}\)). The sketch shows that the shell method is a good choice for solving this problem.

The volume is given by\[V=2 \pi \int ^1_0 x \cdot \dfrac{x^2}{(x^2+1)^2}\,dx = 2 \pi \int ^1_0 \dfrac{x^3}{(x^2+1)^2}\,dx.\nonumber\]Since \(\deg((x^2+1)^2) = 4 \gt 3 = \deg(x^3)\), we can proceed with partial fraction decomposition. Note that \((x^2+1)^2\) is a repeated irreducible quadratic. Using the decomposition described in the problem-solving strategy, we get\[\dfrac{x^3}{(x^2+1)^2} = \dfrac{Ax+B}{x^2+1}+\dfrac{Cx+D}{(x^2+1)^2}.\nonumber\]Multiplying both sides by the LCD gives\[x^3=(Ax+B)(x^2+1)+Cx+D.\nonumber\]Solving, we obtain \(A=1\), \(B=0\), \(C=-1\), and \(D=0\). Substituting back into the integral, we have\[V=2 \pi \int _0^1\dfrac{x^3}{(x^2+1)^2}\,dx=2 \pi \int _0^1\left(\dfrac{x}{x^2+1}-\dfrac{x}{(x^2+1)^2}\right)\,dx=2 \pi \left(\dfrac{1}{2}\ln (x^2+1)+\dfrac{1}{2} \cdot \dfrac{1}{x^2+1}\right)\bigg|^1_0= \pi \left(\ln 2-\tfrac{1}{2}\right).\nonumber\]