Solutions to Problem Set 6

Math 211-03

9-14-2017

[Partial fractions]

1. Compute $\displaystyle \int
   \dfrac{8 x^2 - 59 x + 106}{(x - 4)^2 (x - 3)}\,dx$ .

$$\eqalign{ \dfrac{8 x^2 - 59 x + 106}{(x - 4)^2 (x - 3)} & = \dfrac{a}{x - 4} + \dfrac{b}{(x - 4)^2} + \dfrac{c}{x - 3} \cr \noalign{\vskip2pt} 8 x^2 - 59 x + 106 & = a (x - 4)(x - 3) + b (x - 3) + c (x - 4)^2 \cr}$$

Set $x = 4$ . This gives

$$-2 = b \cdot 1, \quad\hbox{so}\quad b = -2.$$

Set $x = 3$ . This gives

$$1 = c \cdot 1, \quad\hbox{so}\quad c = 1.$$

Plugging the values for b and c back in yields

$$8 x^2 - 59 x + 106 = a (x - 4)(x - 3) - 2 (x - 3) + (x - 4)^2.$$

Set $x = 0$ . This gives

$$106 = a \cdot 12 - -6 + 16, \quad\hbox{so}\quad a = 7.$$

Thus,

$$\dfrac{8 x^2 - 59 x + 106}{(x - 4)^2 (x - 3)} = \dfrac{7}{x - 4} - \dfrac{2}{(x - 4)^2} + \dfrac{1}{x - 3}.$$

Hence,

$$\int \dfrac{8 x^2 - 59 x + 106}{(x - 4)^2 (x - 3)}\,dx = \int \left(\dfrac{7}{x - 4} - \dfrac{2}{(x - 4)^2} + \dfrac{1}{x - 3}\right)\,dx =$$

$$7 \ln |x - 4| + \dfrac{2}{x - 4} + \ln |x - 3| + C.\quad\halmos$$


2. Compute $\displaystyle \int
   \dfrac{5 x^2 - 27 x - 15}{x^3 - 5 x^2}$ .

Note that

$$x^3 - 5 x^2 = x^2 (x - 5).$$

Set up the partial fractions decomposition:

$$\eqalign{ \dfrac{5 x^2 - 27 x - 15}{x^2 (x - 5)} & = \dfrac{a}{x} + \dfrac{b}{x^2} + \dfrac{c}{x - 5} \cr \noalign{\vskip2pt} 5 x^2 - 27 x - 15 & = a x (x - 5) + b (x - 5) + c x^2 \cr}$$

Set $x = 0$ . This gives

$$-15 = -5 b, \quad\hbox{so}\quad b = 3.$$

Set $x = 5$ . This gives

$$-25 = 25 c, \quad\hbox{so}\quad c = -1.$$

Plug $b = 3$ and $c = -1$ back in to get

$$\eqalign{ 5 x^2 - 27 x - 15 & = a x (x - 5) + 3 (x - 5) - x^2 \cr 5 x^2 - 27 x - 15 & = a x (x - 5) + 3 x - 15 - x^2 \cr 6 x^2 - 30 x & = a x (x - 5) \cr 6 x (x - 5) & = a x (x - 5) \cr}$$

From this equation, it follows that $a = 6$ .

Alternatively, you can set x to any number besides 0 or 5 in the first equation of the four above, then solve for a (instead of doing the algebra that I did).

Compute the integral:

$$\int \dfrac{5 x^2 - 27 x - 15}{x^2 (x - 5)} = \int \left(\dfrac{6}{x} + \dfrac{3}{x^2} + \dfrac{-1}{x - 5}\right)\,dx =$$

$$6 \ln |x| - \dfrac{3}{x} - \ln |x - 5| + C.\quad\halmos$$


3. Compute $\displaystyle \int
   \dfrac{5 x^2 + 14 x + 23}{(x + 5)(x^2 + 1)}\,dx$ .

$$\eqalign{ \dfrac{5 x^2 + 14 x + 23}{(x + 5)(x^2 + 1)} & = \dfrac{a}{x + 5} + \dfrac{b x + c}{x^2 + 1} \cr 5 x^2 + 14 x + 23 & = a(x^2 + 1) + (b x + c)(x + 5) \cr}$$

Set $x = -5$ . This gives

$$78 = 26 a, \quad\hbox{so}\quad a = 3.$$

Plugging this back in, I get

$$5 x^2 + 14 x + 23 = 3(x^2 + 1) + (b x + c)(x + 5).$$

Set $x = 0$ . This gives

$$23 = 3 + 5 c, \quad\hbox{so}\quad c = 4.$$

Plugging this back in, I get

$$5 x^2 + 14 x + 23 = 3 (x^2 + 1) + (b x + 4)(x + 5).$$

Now plugging any number in for x (except -5) allows me to solve for b.

For example, set $x = 1$ . Then

$$42 = 6 + (6)(b + 4), \quad\hbox{so}\quad b = 2.$$

With the coefficients found, I can do the integral:

$$\int \left(\dfrac{3}{x + 5} + \dfrac{2 x}{x^2 + 1} + \dfrac{4}{x^2 + 1}\right)\,dx = 3 \ln |x + 5| + \ln |x^2 + 1| + 4 \tan^{-1} x + K.\quad\halmos$$


4. Compute $\displaystyle \int
   \dfrac{2 x^3 + 5 x^2 + 6 x + 5}{(x^2 + 1)^2}\,dx$ .

$$\eqalign{ \dfrac{2 x^3 + 5 x^2 + 6 x + 5}{(x^2 + 1)^2} & = \dfrac{a x + b}{x^2 + 1} + \dfrac{c x + d}{(x^2 + 1)^2} \cr 2 x^3 + 5 x^2 + 6 x + 5 & = (a x + b)(x^2 + 1) + (c x + d) \cr}$$

First, set $x = 0$ in the last x-equation. This gives

$$5 = b + d.$$

Save this equation. Now take the original equation and differentiate with respect to x:

$$6 x^2 + 10 x + 6 = (a x + b)(2 x) + a (x^2 + 1) + c.$$

I used the Product Rule to differentiate $(a x + b)(x^2 + 1)$ .

Set $x = 0$ again. (I'm using $x = 0$ because it's easy to do the computations, but you could use other numbers.) This gives

$$6 = a + c.$$

Save this equation. Differentiate the last x-equation:

$$12 x + 10 = 2(a x + b) + 2 a x + 2 a x.$$

Set $x = 0$ again. This gives

$$10 = 2 b, \quad\hbox{so}\quad b = 5.$$

Plug this into $5 = b + d$ to get

$$5 = 5 + d, \quad\hbox{so}\quad d = 0.$$

Differentiate the last x-equation:

$$12 = 2 a + 2 a + 2 a, \quad\hbox{so}\quad 12 = 6 a, \quad\hbox{and}\quad a = 2.$$

Plug this into $6 = a + c$ to get

$$6 = 2 + c, \quad\hbox{so}\quad c = 4.$$

Now do the integral:

$$\int \left(\dfrac{2 x}{x^2 + 1} + \dfrac{5}{x^2 + 1} + \dfrac{4 x}{(x^2 + 1)^2}\right)\,dx = \ln |x^2 + 1| + 5 \tan^{-1} x - \dfrac{2}{x^2 + 1} + K.$$

I integrated the first and third terms using the substitution $u = x^2 + 1$ .


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