Calculus/Integration techniques/Partial Fraction Decomposition/Solutions

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Evaluate the following by the method partial fraction decomposition.

1. \int\frac{2x+11}{(x+6)(x+5)}dx

Decompose the fraction:

\frac{2x+11}{(x+6)(x+5)}=\frac{A}{x+6}+\frac{B}{x+5}=\frac{Ax+5A+Bx+6B}{(x+6)(x+5}

Equate coefficients of x:

A+B=2
5A+6B=11

Solve the system of equations:

A=1 , B=1

Rewrite the integral and solve:

\begin{align}\int\frac{2x+11}{(x+6)(x+5)}dx&=\int\frac{dx}{x+6}+\int\frac{dx}{x+5}\\
&=\mathbf{\ln|x+6|+\ln|x+5|+C}\end{align}
2. \int\frac{7x^{2}-5x+6}{(x-1)(x-3)(x-7)}dx

Decompose the fraction:

\begin{align}\frac{7x^{2}-5x+6}{(x-1)(x-3)(x-7)}&=\frac{A}{x-1}+\frac{B}{x-3}+\frac{C}{x-7}\\
&=\frac{A(x-3)(x-7)+B(x-1)(x-7)+C(x-1)(x-3)}{(x-1)(x-3)(x-7)}\\
&=\frac{A(x^{2}-10x+21)+B(x^{2}-8x+7)+C(x^{2}-4x+3)}{(x-1)(x-3)(x-7)}\\
&=\frac{Ax^{2}-10Ax+21A+Bx^{2}-8Bx+7B+Cx^{2}-4Cx+3C}{(x-1)(x-3)(x-7)}\end{align}

Equate coefficients:

\begin{align}A+B+C&=7\\
-10A-8B-4C&=-5\\
21A+7B+3C&=6\end{align}

Solve the system of equations:

D=\Biggr|\begin{array}{ccc}
1 & 1 & 1\\
-10 & -8 & -4\\
21 & 7 & 3
\end{array}\Biggr|=-24-84-70-(-30-28-168)=48
A=\frac{\Biggr|\begin{array}{ccc}
7 & 1 & 1\\
-5 & -8 & -4\\
6 & 7 & 3
\end{array}\Biggr|}{D}=\frac{(7)(-8)(3)-24-35-(-15+(7)(-4)(7)-48)}{48}=\frac{32}{48}=\frac{2}{3}
B=\frac{\Biggr|\begin{array}{ccc}
1 & 7 & 1\\
-10 & -5 & -4\\
21 & 6 & 3
\end{array}\Biggr|}{D}=\frac{-15+(7)(-4)(21)-60-(-210-24+(-5)(21))}{48}=\frac{-324}{48}=-\frac{27}{4}
C=\frac{\Biggr|\begin{array}{ccc}
1 & 1 & 7\\
-10 & -8 & -5\\
21 & 7 & 6
\end{array}\Biggr|}{D}=\frac{(-8)(6)+(-5)(21)+(7)(-10)(7)-((-10)(6)+(-5)(7)+(7)(-8)(21))}{48}=\frac{628}{48}=\frac{157}{12}

Rewrite the integral and solve:

\begin{align}\int\frac{7x^{2}-5x+6}{(x-1)(x-3)(x-7)}dx&=\frac{2}{3}\int\frac{dx}{x-1}-\frac{27}{4}\int\frac{dx}{x-3}+\frac{157}{12}\int\frac{dx}{x-7}\\
&=\mathbf{\frac{2}{3}\ln|x-1|-\frac{27}{4}\ln|x-3|+\frac{157}{12}\ln|x-7|+C}\end{align}
3. \int\frac{x^{2}-x+2}{x(x+2)^{2}}dx

Decompose the fraction:

\begin{align}\frac{x^{2}-x+2}{x(x+2)^{2}}&=\frac{A}{x}+\frac{B}{x+2}+\frac{C}{(x+2)^{2}}\\
&=\frac{A(x+2)^{2}+Bx(x+2)+Cx}{x(x+2)^{2}}\\
&=\frac{A(x^{2}+4x+4)+Bx^{2}+2Bx+Cx}{x(x+2)^{2}}\\
&=\frac{Ax^{2}+4Ax+4A+Bx^{2}+2Bx+Cx}{x(x+2)^{2}}\end{align}

Equate the coefficients:

\begin{array}{ccccccc}
A & + & B &  &  & = & 1\\
4A & + & 2B & + & C & = & -1\\
4A &  &  &  &  & = & 2
\end{array}

Solve the system of equations:

4A=2\implies A=\frac{1}{2}
A+B=1\implies B=\frac{1}{2}
4A+2B+C=-1\implies C=-4

Rewrite the integral and solve:

\begin{align}\int\frac{x^{2}-x+2}{x(x+2)^{2}}dx&=\frac{1}{2}\int\frac{dx}{x}+\frac{1}{2}\int\frac{dx}{x+2}-4\int\frac{dx}{(x+2)^{2}}\\
&=\mathbf{\frac{1}{2}\ln|x|+\frac{1}{2}\ln|x+2|+\frac{4}{x+2}+C}\end{align}
4. \int \frac{2}{(x+2)(x^{2}+3)} dx
\frac{2}{(x+2)(x^{2}+3)}=\frac{A}{x+2}+\frac{Bx+C}{x^{2}+3}=\frac{Ax^{2}+3A+Bx^{2}+Cx+2Bx+2C}{(x+2)(x^{2}+3)}

Equate the coefficients for each power of x. For x^2:

A+B=0

, for x:

C+2B=0

, and for the constant terms:

3A+2C=2

Solve the system of equations however you see fit (Gaussian elimination with back-substitution used here):


\left[\begin{array}{cccc}
1 & 1 & 0 & 0\\
0 & 2 & 1 & 0\\
3 & 0 & 2 & 2
\end{array}\right] 

\left[\begin{array}{cccc}
1 & 1 & 0 & 0\\
0 & 1 & \frac{1}{2} & 0\\
0 & -3 & 2 & 2
\end{array}\right] 

\left[\begin{array}{cccc}
1 & 1 & 0 & 0\\
0 & 1 & \frac{1}{2} & 0\\
0 & 0 & \frac{7}{2} & 2
\end{array}\right] 

\left[\begin{array}{cccc}
1 & 1 & 0 & 0\\
0 & 1 & \frac{1}{2} & 0\\
0 & 0 & 1 & \frac{4}{7}
\end{array}\right]
C=\frac{4}{7},\,
B=-\frac{2}{7},\,
A=\frac{2}{7}

So

\int\frac{2}{(x+2)(x^{2}+3)}dx=\frac{2}{7}\int\frac{1}{x+2}dx-\frac{2}{7}\int\frac{x}{x^2+3}+\frac{4}{7}\int\frac{1}{x^{2}+3}dx

To evaluate the first integral use substitution, letting u=x+2, du=dx.
To evaluate the second integral use substitution, letting u=x^{2}+3, du=2xdx, dx=\frac{du}{2x}.
To evaluate the third integral, use the trigonometric substitution x=\sqrt{3}\tan(\theta), dx=\sqrt{3}\sec^{2}(\theta)d\theta.

\begin{align}\int\frac{2}{(x+2)(x^{2}+3)}dx&=\frac{2}{7}\int\frac{1}{x+2}dx-\frac{2}{7}\int\frac{x}{x^2+3}+\frac{4}{7}\int\frac{1}{x^{2}+3}dx\\
&=\frac{2}{7}\ln|x+2|-\frac{2}{14}\ln|x^{2}+3|+\frac{4}{7}\int\frac{\sqrt{3}\sec^{2}(x)d\theta}{3\tan^{2}(x)+3}\\
&=\mathbf{\frac{2}{7}\ln|x+2|-\frac{1}{7}\ln|x^{2}+3|+\frac{4}{7\sqrt{3}}\arctan(\frac{x}{\sqrt{3}})}\end{align}
5. \int\frac{dx}{(x+2)(x^{2}+2)}

Decompose the fraction:

\begin{align}\frac{1}{(x+2)(x^{2}+2)}&=\frac{A}{x+2}+\frac{Bx+C}{x^{2}+2}\\
&=\frac{A(x^{2}+2)+(Bx+C)(x+2)}{(x+2)(x^{2}+2)}\\
&=\frac{Ax^{2}+2A+Bx^{2}+2Bx+Cx+2C}{(x+2)(x^{2}+2)}\end{align}

Equate the coefficients:

\begin{array}{ccccccc}
A & + & B &  &  & = & 0\\
 &  & 2B & + & C & = & 0\\
2A &  &  & + & 2C & = & 1
\end{array}

Solve the system of equations:

D=\Biggr|\begin{array}{ccc}
1 & 1 & 0\\
0 & 2 & 1\\
2 & 0 & 2
\end{array}\Biggr|=6
A=\frac{\Biggr|\begin{array}{ccc}
0 & 1 & 0\\
0 & 2 & 1\\
1 & 0 & 2
\end{array}\Biggr|}{D}=\frac{1}{6}
B=\frac{\Biggr|\begin{array}{ccc}
1 & 0 & 0\\
0 & 0 & 1\\
2 & 1 & 2
\end{array}\Biggr|}{D}=-\frac{1}{6}
C=\frac{\Biggr|\begin{array}{ccc}
1 & 1 & 0\\
0 & 2 & 0\\
2 & 0 & 1
\end{array}\Biggr|}{D}=\frac{2}{6}=\frac{1}{3}

Rewrite the integral and solve:

\int\frac{dx}{(x+2)(x^{2}+2)}=\frac{1}{6}\int\frac{dx}{x+2}-\frac{1}{6}\int\frac{xdx}{x^{2}+2}+\frac{1}{3}\int\frac{dx}{x^{2}+2}

Making the substitution

u=x^{2}+2\qquad du=2xdx\qquad dx=\frac{du}{2x}

in the second integral and

\frac{x}{\sqrt{2}}=\tan(\theta)\qquad\frac{dx}{\sqrt{2}}=\sec^{2}(\theta)d\theta\qquad dx=\sqrt{2}\sec^{2}(\theta)d\theta

in the third integral, we have

\begin{align}\int\frac{dx}{(x+2)(x^{2}+2)}&=\frac{1}{6}\ln|x+2|-\frac{1}{6}\int\frac{du}{2u}+\frac{1}{3}\int\frac{\sqrt{2}\sec^{2}(\theta)d\theta}{2(\tan^{2}(\theta)+1)}\\
&=\frac{1}{6}\ln|x+2|-\frac{1}{12}\ln|u|+\frac{\sqrt{2}}{6}\int\frac{\sec^{2}(\theta)d\theta}{\sec^{2}(\theta)}\\
&=\frac{1}{6}\ln|x+2|-\frac{1}{12}\ln|x^{2}+2|+\frac{\sqrt{2}}{6}\int d\theta\\
&=\frac{1}{6}\ln|x+2|-\frac{1}{12}\ln|x^{2}+2|+\frac{\sqrt{2}}{6}\theta+C\\
&=\mathbf{\frac{1}{6}\ln|x+2|-\frac{1}{12}\ln|x^{2}+2|+\frac{\sqrt{2}}{6}\arctan(\frac{x}{\sqrt{2}})+C}\end{align}
6. \int\frac{dx}{(x^{2}+1)^{2}(x-1)}

Decompose the fraction:

\begin{align}\frac{1}{(x^{2}+1)^{2}(x-1)}&=\frac{Ax+B}{x^{2}+1}+\frac{Cx+D}{(x^{2}+1)^{2}}+\frac{E}{x-1}\\
&=\frac{(Ax+B)(x^{2}+1)(x-1)+(Cx+D)(x-1)+E(x^{2}+1)^{2}}{(x^{2}+1)^{2}(x-1)}\\
&=\frac{(Ax+B)(x^{3}-x^{2}+x-1)+Cx^{2}+(D-C)x-D+E(x^{4}+2x^{2}+1)}{(x^{2}+1)^{2}(x-1)}\\
&=\frac{Ax^{4}+(B-A)x^{3}+(A-B)x^{2}+(B-A)x-B+Cx^{2}+(D-C)x-D+Ex^{4}+2Ex^{2}+E}{(x^{2}+1)^{2}(x-1)}\end{align}

Equate coefficients:

\begin{array}{ccccccccccc}
A &  &  &  &  &  &  & + & E & = & 0\\
-A & + & B &  &  &  &  &  &  & = & 0\\
A & - & B & + & C &  &  & + & 2E & = & 0\\
-A & + & B & - & C & + & D &  &  & = & 0\\
 & - & B &  &  & - & D & + & E & = & 0
\end{array}

Solve the system of equations any way you see fit. Here, we'll solve for A by Cramer's rule, then plug in to solve for the other variables. The denominator in Cramer's rule will be

d=\left|\begin{array}{ccccc}
1 & 0 & 0 & 0 & 1\\
-1 & 1 & 0 & 0 & 0\\
1 & -1 & 1 & 0 & 2\\
-1 & 1 & -1 & 1 & 0\\
0 & -1 & 0 & -1 & 1
\end{array}\right|

Expanding across the top row gives

d=\left|\begin{array}{cccc}
1 & 0 & 0 & 0\\
-1 & 1 & 0 & 2\\
1 & -1 & 1 & 0\\
-1 & 0 & -1 & 1
\end{array}\right|+\left|\begin{array}{cccc}
-1 & 1 & 0 & 0\\
1 & -1 & 1 & 0\\
-1 & 1 & -1 & 1\\
0 & -1 & 0 & -1
\end{array}\right|

Expanding across the top rows in both matrices gives

d=\left|\begin{array}{ccc}
1 & 0 & 2\\
-1 & 1 & 0\\
0 & -1 & 1
\end{array}\right|-\left|\begin{array}{ccc}
-1 & 1 & 0\\
1 & -1 & 1\\
-1 & 0 & -1
\end{array}\right|-\left|\begin{array}{ccc}
1 & 1 & 0\\
-1 & -1 & 1\\
0 & 0 & -1
\end{array}\right|

Solve the individual determinants

\left|\begin{array}{ccc}
1 & 0 & 2\\
-1 & 1 & 0\\
0 & -1 & 1
\end{array}\right|=(1)(1)(1)+(0)(0)(1)+(2)(-1)(-1)-(0)(-1)(2)-(1)(0)(-1)-(2)(1)(0)=1+2=3
\left|\begin{array}{ccc}
-1 & 1 & 0\\
1 & -1 & 1\\
-1 & 0 & -1
\end{array}\right|=(-1)(-1)(-1)+(1)(1)(-1)+(0)(1)(0)-(1)(1)(-1)-(-1)(1)(0)-(0)(-1)(-1)=-1-1-(-1)=-1
\left|\begin{array}{ccc}
1 & 1 & 0\\
-1 & -1 & 1\\
0 & 0 & -1
\end{array}\right|\begin{array}{cc}
1 & 1\\
-1 & -1\\
0 & 0
\end{array}=(1)(-1)(-1)+(1)(1)(0)+(0)(-1)(0)-(1)(-1)(-1)-(1)(1)(0)-(0)(-1)(0)=1-1=0

So

d=3-(-1)-0=4

Now use Cramer's rule to solve for A :

A=\frac{1}{4}\left|\begin{array}{ccccc}
0 & 0 & 0 & 0 & 1\\
0 & 1 & 0 & 0 & 0\\
0 & -1 & 1 & 0 & 2\\
0 & 1 & -1 & 1 & 0\\
1 & -1 & 0 & -1 & 1
\end{array}\right|

Expanding down the first column gives

A=\frac{1}{4}\left|\begin{array}{cccc}
0 & 0 & 0 & 1\\
1 & 0 & 0 & 0\\
-1 & 1 & 0 & 2\\
1 & -1 & 1 & 0
\end{array}\right|

Expanding across the first row gives

A=-\frac{1}{4}\left|\begin{array}{ccc}
1 & 0 & 0\\
-1 & 1 & 0\\
1 & -1 & 1
\end{array}\right|

Expanding down the last column gives

A=-\frac{1}{4}\left|\begin{array}{cc}
1 & 0\\
-1 & 1
\end{array}\right|=-\frac{1}{4}

Now that we know A, we can solve for E using the first equation

-\frac{1}{4}+E=0\implies E=\frac{1}{4}

We can solve for B using the second equation and the value of A

-(-\frac{1}{4})+B=0\implies B=-\frac{1}{4}

We can solve for C using the third equation and the values we've found so far

-\frac{1}{4}-(-\frac{1}{4})+C+2\frac{1}{4}=0\implies C=-\frac{1}{2}

We can solve for D using the last equation and the values of B and E

-(-\frac{1}{4})-D+\frac{1}{4}=1\implies D=-\frac{1}{2}

Finally, we can check our solution using the 4th equation and the values we've found

-(-\frac{1}{4})+(-\frac{1}{4})-(-\frac{1}{2})+(-\frac{1}{2})=0\checkmark

Rewrite the integral and solve

\int\frac{dx}{(x^{2}+1)^{2}(x-1)}=-\frac{1}{4}\int\frac{xdx}{x^{2}+1}-\frac{1}{4}\int\frac{dx}{x^{2}+1}-\frac{1}{2}\int\frac{xdx}{(x^{2}+1)^{2}}-\frac{1}{2}\int\frac{dx}{(x^{2}+1)^{2}}+\frac{1}{4}\int\frac{dx}{x-1}

Let's solve each integral separately. To solve the first, use the substitution

u=x^{2}+1;\qquad du=2xdx;\qquad dx=\frac{du}{2x}
\int\frac{xdx}{x^{2}+1}=\int\frac{du}{2u}=\frac{1}{2}\ln|u|+C_{1}=\frac{1}{2}\ln|x^{2}+1|+C_{1}

To solve the second integral, use the substitution

x=\tan(\theta);\qquad dx=\sec^{2}(\theta)d\theta
\int\frac{dx}{x^{2}+1}=\int\frac{\sec^{2}(\theta)d\theta}{\tan^{2}(\theta)+1}=\int\frac{\sec^{2}(\theta)d\theta}{\sec^{2}(\theta)}=\int d\theta=\theta+C_{2}=\arctan(x)+C_{2}

To solve the third integral, use the substitution

u=x^{2}+1;\qquad du=2xdx;\qquad dx=\frac{du}{2x}
\int\frac{xdx}{(x^{2}+1)^{2}}=\int\frac{du}{2u^{2}}=-\frac{1}{2u}+C_{3}=-\frac{1}{2(x^{2}+1)}+C_{3}

To solve the fourth integral, use the substitution

x=\tan(\theta);\qquad dx=\sec^{2}(\theta)d\theta
\begin{align}\int\frac{dx}{(x^{2}+1)^{2}}&=\int\frac{\sec^{2}(\theta)d\theta}{(\tan^{2}(\theta)+1)^{2}}\\
&=\int\frac{\sec^{2}(\theta)d\theta}{(\sec^{2}(\theta))^{2}}\\
&=\int\frac{d\theta}{\sec^{2}(\theta)}\\
&=\int\cos^{2}(\theta)d\theta\\
&=\int\frac{1+\cos(2\theta)}{2}d\theta\\
&=\int\frac{d\theta}{2}+\frac{1}{2}\int\cos(2\theta)d\theta\\
&=\frac{\theta}{2}+\frac{1}{4}\sin(2\theta)+C_{4}\\
&=\frac{\theta}{2}+\frac{1}{2}\cos(\theta)\sin(\theta)+C_{4}\\
&=\frac{1}{2}\arctan(x)+\frac{1}{2}\cos(\theta)\sin(\theta)+C_{4}\\
&=\frac{1}{2}\arctan(x)+\frac{1}{2}\frac{1}{\sqrt{1+x^{2}}}\frac{x}{\sqrt{1+x^{2}}}+C_{4}\\
&=\frac{1}{2}\arctan(x)+\frac{x}{2(1+x^{2})}+C_{4}\end{align}

To solve the last integral, use the substitution

u=x-1;\qquad du=dx
\int\frac{dx}{x-1}=\int\frac{du}{u}=\ln|u|+C_{5}=\ln|x-1|+C_{5}

Putting it all together, we have

\begin{align}\int\frac{dx}{(x^{2}+1)^{2}(x-1)}&=-\frac{1}{4}(\frac{1}{2}\ln|x^{2}+1|)-\frac{1}{4}\arctan(x)-\frac{1}{2}(-\frac{1}{2(x^{2}+1)})-\frac{1}{2}(\frac{1}{2}\arctan(x)+\frac{x}{2(1+x^{2})})+\frac{1}{4}\ln|x-1|+C\\
&=-\frac{1}{8}\ln|x^{2}+1|-\frac{1}{4}\arctan(x)+\frac{1}{4(x^{2}+1)}-\frac{1}{4}\arctan(x)-\frac{x}{4(1+x^{2})}+\frac{1}{4}\ln|x-1|+C\\
&=\mathbf{-\frac{1}{2}\arctan(x)+\frac{1-x}{4(x^{2}+1)}+\frac{1}{8}\ln\left(\frac{(x-1)^{2}}{x^{2}+1}\right)+C}\end{align}