Solutions to Problem Set 7

Math 211-03

9-18-2017

[Miscellaneous substitutions]

1. Compute $\displaystyle \int
   \dfrac{1}{x^2 + 10 x + 169}\,dx$ .

The quadratic $x^2 + 10 x + 169$ does not factor. The middle term in the quadratic is $10 x$ . Since $\dfrac{1}{2} \cdot 10 = 5$ and $5^2
   = 25$ , I have

$$x^2 + 10 x + 169 = x^2 + 10 x + 25 + 144 = (x + 5)^2 + 144.$$

Therefore,

$$\int \dfrac{1}{x^2 + 10 x + 169}\,dx = \int \dfrac{1}{(x + 5)^2 + 144}\,dx = \int \dfrac{1}{u^2 + 144}\,du = \dfrac{1}{12} \tan^{-1} \dfrac{u}{12} + C = \dfrac{1}{12} \tan^{-1} \dfrac{x + 5}{12} + C.$$

$$\left[u = x + 5, \quad du = dx\right]\quad\halmos$$


2. Compute $\displaystyle \int
   \dfrac{1}{x^2 - 4 x - 5}\,dx$ .

Since $x^2 - 4 x - 5$ factors into $(x - 5)(x + 1)$ , I'll use partial fractions. (You could do this by completing the square, then doing a trig substitution. I think it would be a little longer.)

Write

$$\eqalign{ \dfrac{1}{x^2 - 4 x - 5} & = \dfrac{a}{x - 5} + \dfrac{b}{x + 1} \cr \noalign{\vskip2pt} 1 & = a (x + 1) + b (x - 5) \cr}$$

Let $x = -1$ . I get $1 =
   -6b$ , so $b = -\dfrac{1}{6}$ .

Let $x = 5$ . I get $1 = 6a$ , so $a = \dfrac{1}{6}$ .

Therefore,

$$\int \dfrac{1}{x^2 - 4 x - 5}\,dx = \int \left(\dfrac{1}{6} \cdot \dfrac{1}{x - 5} - \dfrac{1}{6} \cdot \dfrac{1}{x + 1}\right)\,dx = \dfrac{1}{6} \ln |x - 5| - \dfrac{1}{6} \ln |x + 1| + C. \quad\halmos$$


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

While $x^2 - 6 x + 5 = (x - 1)(x
   - 5)$ , factoring doesn't help. I'll complete the square. The middle term is $6 x$ . Since $\dfrac{1}{2}\cdot 6 = 3$ and $3^2 = 9$ , I have

$$-x^2 + 6 x - 5 = -(x^2 - 6 x) - 5 = -(x^2 - 6 x + 9) + (9 - 5) = 4 - (x - 3)^2.$$

Therefore,

$$\int \dfrac{1}{(-x^2 + 6 x - 5)^{3/2}}\,dx = \int \dfrac{1}{\left[4 - (x - 3)^2\right]^{3/2}}\,dx = \int \dfrac{1}{\left[4 - 4 (\sin \theta)^2\right]^{3/2}} (2 \cos \theta\,d\theta) =$$

$$\hfil\raise0.5in\hbox{$\left[x - 3 = 2 \sin \theta, \quad dx = 2 \cos \theta\,d\theta\right]$} \hskip0.5in \hbox{\epsfysize=1in \epsffile{miscellaneous-substitutions3.eps}}\hfil$$

$$\int \dfrac{1}{\left[4 (\cos \theta)^2\right]^{3/2}} (2 \cos \theta\,d\theta) = \int \dfrac{2 \cos \theta}{8 (\cos \theta)^3}\,d\theta = \dfrac{1}{4} \int \dfrac{1}{(\cos \theta)^2}\,d\theta = \dfrac{1}{4} \int (\sec \theta)^2\,d\theta =$$

$$\dfrac{1}{4} \tan \theta + C = \dfrac{1}{4} \dfrac{x - 3}{\sqrt{-x^2 + 6 x - 5}} + C.\quad\halmos$$


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

The quadratic $x^2 + 2 x + 2$ does not factor. I'll use partial fractions.

$$\eqalign{ \dfrac{2 x^2 + 5 x + 3}{(x - 1)(x^2 + 2 x + 2)} & = \dfrac{a}{x - 1} + \dfrac{b x + c}{x^2 + 2 x + 2} \cr \noalign{\vskip2pt} 2 x^2 + 5 x + 3 & = a (x^2 + 2 x + 2) + (b x + c)(x - 1) \cr}$$

Let $x = 1$ . I get $10 = 5
   a$ , so $a = 2$ . Then

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

Let $x = 0$ . I get $3 = 4 -
   c$ , so $c = 1$ . Then

$$2 x^2 + 5 x + 3 = 2 (x^2 + 2 x + 2) + (b x + 1)(x - 1).$$

At this point, I can reduce the equation by differentiating both sides, or I can plug another value in for x. I'll let $x = 2$ . I get

$$21 = 20 + (2 b + 1)(1), \quad\hbox{so}\quad b = 0.$$

Therefore,

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

For the first term, I have

$$\int \dfrac{2}{x - 1}\,dx = 2 \ln |x - 1| + C.$$

For the second term, I complete the square:

$$\int \dfrac{1}{x^2 + 2 x + 2}\,dx = \int \dfrac{1}{1 + (x + 1)^2}\,dx = \tan^{-1} (x + 1) + C.$$

(The last step follows by making the easy substitution $u = x + 1$ . You can work out the details.)

Therefore, the answer to the original question is

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


5. Compute $\displaystyle \int
   \dfrac{1}{x - x^{2/3}}\,dx$ .

$$\int \dfrac{1}{x - x^{2/3}}\,dx = \int \dfrac{3 u^{2}}{u^{3} - u^{2}}\,du = 3 \int \dfrac{1}{u - 1}\,du =$$

$$\left[x = u^{3}, \quad dx = 3 u^{2}\,du\right]$$

$$3 \ln |u - 1| + C = 3 \ln |x^{1/3} - 1| + C.\quad\halmos$$


6. Compute $\displaystyle \int
   \dfrac{10}{x^{3/2} - 4 x^{1/2}}\,dx$ .

Since I see lots of powers of $x^{1/2}$ , I'll use $x = u^2$ :

$$\int \dfrac{10}{x^{3/2} - 4 x^{1/2}}\,dx = 10 \int \dfrac{1}{(u^2)^{3/2} - 4 (u^2)^{1/2}}(2 u\,du) = 20 \int \dfrac{u}{u^3 - 4 u}\,du =$$

$$\left[x = u^2, \quad dx = 2 u\,du; \quad u = x^{1/2}\right]$$

$$20 \int \dfrac{1}{u^2 - 4}\,du = \int \dfrac{20}{(u - 2)(u + 2)}\,du.$$

Apply partial fractions:

$$\eqalign{ \dfrac{20}{(u - 2)(u + 2)} & = \dfrac{a}{u - 2} + \dfrac{b}{u + 2} \cr \noalign{\vskip2pt} 20 & = a (u + 2) + b (u - 2) \cr}$$

Let $u = 2$ . I get $20 = 4
   a$ , so $a = 5$ .

Let $u = -2$ . I get $20 = -4
   b$ , so $b = -5$ .

Therefore,

$$\int \dfrac{20}{(u - 2)(u + 2)}\,du = \int \left(\dfrac{5}{u - 2} - \dfrac{5}{u + 2}\right)\,du = 5 \ln |u - 2| - 5 \ln |u + 2| + C =$$

$$5 \ln |x^{1/2} - 2| - 5 \ln |x^{1/2} + 2| + C.\quad\halmos$$


7. Compute $\displaystyle \int
   \dfrac{1}{x^{1/2} - x^{1/3}}\,dx$ .

$$\int \dfrac{1}{x^{1/2} - x^{1/3}}\,dx = \int \dfrac{1}{(u^6)^{1/2} - (u^6)^{1/3}} \cdot 6 u^5\,du = 6 \int \dfrac{u^5}{u^3 - u^2} \cdot u^5\,du = 6 \int \dfrac{u^5}{u^2 (u - 1)}\,du =$$

$$\left[x = u^6, \quad dx = 6 u^5\,du; \quad u = x^{1/6}\right]$$

$$6 \int \dfrac{u^3}{u - 1}\,du = 6 \int \left(u^2 + u + 1 + \dfrac{1}{u - 1}\right)\,du = 6 \left(\dfrac{1}{3} u^3 + \dfrac{1}{2} u^2 + u + \ln |u - 1|\right) + c =$$

$$6 \left(\dfrac{1}{3} x^{1/2} + \dfrac{1}{2} x^{1/3} + x^{1/6} + \ln |x^{1/6} - 1|\right) + c.$$

Here's the work for the long division:

$$\hbox{\epsfysize=1.5in \epsffile{miscellaneous-substitutions7.eps}}\quad\halmos$$


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