Solutions to Problem Set 8

Math 211-03

9-19-2017

[Integration techniques]

1. Compute $\displaystyle \int
   \dfrac{x^2 + 9 x - 6}{x^2(x - 1)}\,dx$ .

$$\eqalign{ \dfrac{x^2 + 9 x - 6}{x^2(x - 1)} & = \dfrac{a}{x} + \dfrac{b}{x^2} + \dfrac{c}{x - 1} \cr x^2 + 9 x - 6 & = a x(x - 1) + b(x - 1) + c x^2 \cr}$$

Let $x = 0$ . I get $-6 = -b$ , so $b = 6$ .

Let $x = 1$ . I get $4 = c$ .

Then

$$x^2 + 9 x - 6 = a x(x - 1) + 6(x - 1) + 4 x^2.$$

Let $x = 2$ . I get $16 = 2 a
   + 6 + 16$ , so $a = -3$ .

So

$$\int \dfrac{x^2 + 9 x - 6}{x^2(x - 1)}\,dx = \int \left(\dfrac{-3}{x} + \dfrac{6}{x^2} + \dfrac{4}{x - 1}\right)\,dx = -3 \ln |x| - \dfrac{6}{x} + 4 \ln |x - 1| + K.\quad\halmos$$


2. Compute $\displaystyle \int
   \dfrac{1}{(x^2 + 10 x + 26)^{3/2}}\,dx$ .

Notice that

$$x^2 + 10 x + 26 = x^2 + 10 x + 25 + 1 = (x + 5)^2 + 1.$$

So

$$\int \dfrac{1}{(x^2 + 10 x + 26)^{3/2}}\,dx = \int \dfrac{1}{[(x + 5)^2 + 1]^{3/2}}\,dx = \int \dfrac{1}{[(\tan \theta)^2 + 1]^{3/2}} (\sec \theta)^2\,d\theta =$$

$$\hfil\raise0.5in\hbox{$\left[x + 5 = \tan \theta, \quad dx = (\sec \theta)^2\,d\theta\right]$} \hskip0.5in \hbox{\epsfysize=1in \epsffile{integration-techniques2.eps}}\hfil$$

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

$$\sin \theta + c = \dfrac{x + 5}{\sqrt{x^2 + 10 x + 26}} + c.\quad\halmos$$


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

$$\int \dfrac{1}{x^{3/4} (x^{1/4} + x^{1/6})}\,dx = \int \dfrac{1}{x + x^{11/12}}\,dx = \int \dfrac{1}{u^{12} + u^{11}} \cdot12 u^{11}\,du = 12 \int \dfrac{1}{u + 1}\,du =$$

$$\left[x = u^{12}, \quad dx = 12 u^{11}\,du; \quad u = x^{1/12}\right]$$

$$12 \ln |u + 1| + C = 12 \ln |x^{1/12} + 1| + c.\quad\halmos$$


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

$$\int \int \dfrac{1}{x + x^{1/5}}\,dx = \int \dfrac{1}{u^5 + (u^5)^{1/5}} \cdot 5 u^4\,du = 5 \int \dfrac{u^4}{u^5 + u}\,du =$$

$$\left[x = u^5, \quad dx = 5 u^4\,du; \quad u = x^{1/5}\right]$$

$$5 \int \dfrac{u^3}{u^4 + 1}\,du = 5 \int \dfrac{u^3}{w} \cdot \dfrac{dw}{4 u^3} = \dfrac{5}{4} \int \dfrac{1}{w}\,dw = \dfrac{5}{4} \ln |w| + C =$$

$$\left[w = u^4 + 1, \quad dw = 4 u^3\,du, \quad du = \dfrac{dw}{4 u^3}\right]$$

$$\dfrac{5}{4} \ln |u^4 + 1| + C = \dfrac{5}{4} \ln |x^{4/5} + 1| + c.\quad\halmos$$


5. Compute $\displaystyle \int
   \dfrac{1}{\sqrt{x^2 - 25}}\,dx$ .

$$\int \dfrac{1}{\sqrt{x^2 - 25}}\,dx = \int \dfrac{5 \sec \theta \tan \theta}{\sqrt{25 (\sec \theta)^2 - 25}}\,d\theta = \int \dfrac{5 \sec \theta \tan \theta}{\sqrt{25 (\tan \theta)^2}}\,d\theta = \int \dfrac{5 \sec \theta \tan \theta}{5 \tan \theta}\,d\theta =$$

$$\hfil\raise0.5in\hbox{$\left[x = 5 \sec \theta, \quad dx = 5 \sec \theta \tan \theta\,d\theta\right]$} \hskip0.5in \hbox{\epsfysize=1in \epsffile{integration-techniques5.eps}}\hfil$$

$$\int \sec \theta\,d\theta = \ln |\sec \theta + \tan \theta| + c = \ln \left|\dfrac{x}{5} + \dfrac{\sqrt{x^2 - 25}}{5}\right| + c.\quad\halmos$$


6. Compute $\displaystyle \int
   \dfrac{x}{\sqrt{x^2 + 4}}\,dx$ .

$$\int \dfrac{x}{\sqrt{x^2 + 4}}\,dx = \int \dfrac{x}{\sqrt{u}} \cdot \dfrac{du}{2 x} = \dfrac{1}{2} \int \dfrac{1}{\sqrt{u}}\,du = \dfrac{1}{2} \cdot 2 \sqrt{u} + C = \sqrt{x^2 + 4} + c.\quad\halmos$$

$$\left[u = x^2 + 4, \quad du = 2 x\,dx, \quad dx = \dfrac{du}{2 x}\right]$$


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

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

Let $x = 0$ . I get

$$4 = b + d.$$

Differentiate the last x-equation:

$$2(4 x - 1) = (a x + b)(2 x) + (a)(x^2 + 1) + c.$$

Let $x = 0$ . I get

$$-2 = a + c.$$

Differentiate the last x-equation:

$$8 = (a x + b)(2) + (a)(2 x) + (a)(2 x).$$

Let $x = 0$ . I get

$$8 = 2 b, \quad\hbox{so}\quad b = 4.$$

Plugging this into $4 = b + d$ gives

$$4 = 4 + d, \quad\hbox{so}\quad d = 0.$$

Differentiate the last x-equation:

$$0 = 2 a + 2 a + 2 a, \quad\hbox{so}\quad 0 = 6 a, \quad\hbox{or}\quad a = 0.$$

Plug $a = 0$ into $-2 = a +
   c$ to get $c = -2$ .

Therefore,

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

Here's the work for the second term in the integral:

$$\int \dfrac{2 x}{(x^2 + 1)^2}\,dx = \int \dfrac{2 x}{u^2} \cdot \dfrac{du}{2 x} = \int \dfrac{du}{u^2} = -\dfrac{1}{u} + K = -\dfrac{1}{x^2 + 1} + K.$$

$$\left[u = x^2 + 1, \quad du = 2 x\,dx, \quad dx = \dfrac{du}{2 x}\right] \quad\halmos$$


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