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Next: Integration of trigonometric functions Up: INTEGRATION Previous: Iintegration of rational functions

3.4 Answer

3.4

1.

(a)

$\displaystyle{\frac{7}{(x-2)(x+5)}}$ is rational function and the degree of the numerator $<$ the degree of the denominator. So, we use the partial fraction expansion. Then

$\displaystyle \frac{7}{(x-2)(x+5)} = \frac{A}{x-2} + \frac{B}{x+5}$

Now get rid of the denominator.

$\displaystyle 7 = A(x+5) + B(x-2) = (A+B)x + 5A - 2B $

Note that the left and right sides are equal here. We obtain the following equation．

$\displaystyle \left\{\begin{array}{l} A+B = 0\\ 5A- 2B = 7 \end{array}\right.$

Solve this using Cramer's rule. Then we have

$\displaystyle A = \frac{\left\vert \begin{array}{ll} 0 & 1\\ 7 & -2 \end{ar... ...begin{array}{ll} 1 & 1\\ 5 & -2 \end{array}\right\vert} = \frac{-7}{-7} = 1$

Putting these back into the equation above, we have $B = -1$. Then

$\displaystyle \frac{7}{(x-2)(x+5)} = \frac{1}{x-2} + \frac{-1}{x+5}$

Thus, $\displaystyle{\int \frac{7}{(x-2)(x+5)}dx = \int (\frac{1}{x-2} + \frac{-1}{x+5}) dx = \log{\vert x-2\vert} - \log{\vert x+5\vert} + c}$

(b) $\displaystyle{\frac{x^2 + 1}{x(x^2 - 1)}}$ is a rational function with the degree of the numerato $<$ the degree of the denominator. Then using the partial fraction expansion, we have

$\displaystyle \frac{x^2 + 1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x + 1}$

Then

$\displaystyle x^2 + 1 = A(x^2 -1) + B(x^2 + x) + C(x^2 - x)= (A+B+C)x^2 + (B-C)x - A $

Note that the left and right sides are equal here. We obtain the following equation．

$\displaystyle \left\{\begin{array}{l} A+B+C = 1\\ B-C = 0\\ A = -1 \end{array}\right.$

$A = -1$ implies that

$\displaystyle \left\{\begin{array}{l} B+C = 2\\ B-C = 0\\ \end{array}\right.$

Solve this equation uinsg Cramer's rule, we have

$\displaystyle B = \frac{\left\vert \begin{array}{ll} 2 & 1\\ 0 & -1 \end{a... ...egin{array}{ll} 1 & 1 \\ 1 & -1 \end{array}\right\vert} = \frac{-2}{-2} = 1$

Put these back into the above equation. Then we have $C = 1$. Thus,

$\displaystyle \frac{x^2 + 1}{x(x-1)(x+1)} = \frac{-1}{x} + \frac{1}{x-1} + \frac{1}{x + 1}$

Therefore,


$\displaystyle \int{\frac{x^2 + 1}{x(x^2 - 1)}\ dx}$ $\displaystyle =$ $\displaystyle \int(\frac{-1}{x} + \frac{1}{x-1} + \frac{1}{x + 1}) \ dx$  
  $\displaystyle =$ $\displaystyle -\log{\vert x\vert} + \log\vert x-1\vert + \log\vert x+1\vert + c = -\log{\vert x\vert} + \log\vert x^2 -1\vert + c$  

(c) $\displaystyle{\frac{x^2 + 3}{x^2 -3x + 2}}$ is a rational function with the degree of the numerator is greater than the degree of the denominator. So, divide the numerator by the denominator. Then

$\displaystyle \frac{x^2 + 3}{x^2 -3x + 2} = 1 + \frac{3x + 1}{x^2 - 3x + 2}$

Now using the partial fraction expansion, we can write $\frac{x^2 + 3}{x^2 -3x + 2}$ as

$\displaystyle \frac{3x+1}{(x-2)(x-1)} = \frac{A}{x-2} + \frac{B}{x-1} $

Multiply $(x-2)(x-1)$ to both sides. Then

$\displaystyle 3x+ 1 = A(x-1) + B(x-2) = (A+B)x - (A +2B)$

Note that the left and right sides are equal here. We obtain the following equation.

$\displaystyle \left\{\begin{array}{l} A+B = 3\\ A+2B = -1 \end{array}\right.$

Solving this equation using Cramer's rule, we have

$\displaystyle A = \frac{\left\vert \begin{array}{ll} 3 & 1\\ -1 & 2 \end{a... ... \begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right\vert} = \frac{7}{1} = 7$

Put these back into the above equation, we have $B = -4$. Then

$\displaystyle \frac{3x + 1}{x^2 -3x + 2} = \frac{7}{x-2} + \frac{-4}{x-1} $

Thus,


$\displaystyle \int{\frac{x^2 + 3}{x^2 -3x + 2}\ dx}$ $\displaystyle =$ $\displaystyle \int(1 + \frac{7}{x-2} + \frac{-4}{x-1}) \ dx$  
  $\displaystyle =$ $\displaystyle x + 7\log{\vert x-2\vert} -4 \log\vert x-1\vert + c$  

(d) $\displaystyle{\int{\frac{x^2}{(x - 1)^2(x + 1)}\ dx}}$ is a rational function with the degree of the numerator is less than the degree of the denominator. So, using the partial fraction expansion, we have

$\displaystyle \frac{x^2}{(x - 1)^2(x + 1)} = \frac{A}{x-1} + \frac{B}{(x-1)^{2}} + \frac{C}{x+1} $

Simplifying,

$\displaystyle x^2 = A(x-1)(x+1) + B(x+1) + C(x-1)^{2}= (A + C)x^2 + (B-2C)x + (-A + B + C)$

Note that the left and right sides are equal here. We obtain the following equation.

$\displaystyle \left\{\begin{array}{l} A+C = 1\\ B-2C = 0\\ -A + B + C = 0 \end{array}\right.$

Solving this equation using Cramer's rule, we have

$\displaystyle A = \frac{\left\vert \begin{array}{lll} 1 & 0 & 1\\ 0 & 1 & -2... ... 1 & 0 &1 \\ 0 & 1 & -2\\ -1 & 1 & 1 \end{array}\right\vert} = \frac{3}{4}$

Put this back into $A + C = 0$. Then $C = \frac{1}{4}$．Furthermore， $A = \frac{3}{4}, C = \frac{1}{4}$ implies $B = \frac{1}{2}$．Thus,

$\displaystyle \frac{x^2}{(x - 1)^2(x + 1)} = \frac{3}{4(x-1)} + \frac{1}{2(x-1)^{2}} + \frac{1}{4(x+1)} $

Therefore,


$\displaystyle \int{\frac{x^2}{(x - 1)^2(x + 1)}\ dx}$ $\displaystyle =$ $\displaystyle \int(\frac{3}{4(x-1)} + \frac{1}{2(x-1)^{2}} + \frac{1}{4(x+1)}) \ dx$  
  $\displaystyle =$ $\displaystyle \frac{3}{4}\log\vert x-1\vert - \frac{1}{2}(x-1)^{-1} + \frac{1}{4}\log\vert x+1\vert + c$  

(e) $\displaystyle{\int{\frac{dx}{(x^2 + 16)^2}}}$ is a rational function with the degree of the numerator is less than the degree of the denominator. So, we might use the partial fraction expansion. But the numerator is already constant. It means that no more partial fraction expansion is necessary. In fact,

$\displaystyle \frac{dx}{(x^2 + 16)^2} = \frac{Ax + B}{x^{2} + 16} + \frac{Cx +D}{(x^2 + 16)^{2}} $

Simplifying, we have

$\displaystyle 1 = (Ax + B)(x^2 + 16) + (Cx+D)= Ax^3 + Bx^2 + (16A + C)x + 16B+D$

Note that the left and right sides are equal here. We obtain the following equation.

$\displaystyle \left\{\begin{array}{l} A = 0\\ B = 0\\ 16A + C = 0\\ 16B + D = 1 \end{array}\right.$

Thus we have $A = B= C =0, D = 1$ which is the original problem.

So, let $\displaystyle{I_{n} = \int{\frac{dx}{(x^2 + 16)^n}}}$ and use the integration by parts.

\begin{displaymath}\begin{array}{ll} u = \frac{1}{(x^2 + 16)}, & dv = dx\\ du = -\frac{2x}{(x^2 + 16)^2}dx, & v = x \end{array}\end{displaymath}

より
$\displaystyle I_{1}$ $\displaystyle =$ $\displaystyle \int{\frac{dx}{(x^2 + 16)}} = \frac{x}{(x^2 + 16)} + 2\int{\frac{x^2}{(x^2 + 16)^2}}dx$  
  $\displaystyle =$ $\displaystyle \frac{x}{(x^2 + 16)} + 2\int{\frac{x^2+16 -16}{(x^2 + 16)^2}}dx$  
  $\displaystyle =$ $\displaystyle \frac{x}{(x^2 + 16)} + 2\int{\frac{1}{(x^2 + 16)^2}}dx -32 \int{\frac{1}{(x^2 + 16)^2}}dx$  
  $\displaystyle =$ $\displaystyle \frac{x}{(x^2 + 16)} + 2I_{1} - 32I_{2}$  

Thus，

$\displaystyle I_{2} = \frac{1}{32}[\frac{x}{(x^2 + 16)} + I_{1}] = \frac{1}{128}[\frac{4x}{(x^2 + 16)} + \tan^{-1}(\frac{x}{4})] + c$

Alternate solution (trig integral) Let $x = 4\tan{\theta}$. Then

$\displaystyle \left\{\begin{array}{ll} x &= 4\tan{\theta}\\ dx &= 4\sec^{2}{\theta}d\theta\\ \sqrt{x^2 + 16} &= 4\sec{\theta} \end{array}\right.$

Thus, we have
$\displaystyle \int{\frac{dx}{(x^2 + 16)^2}}$ $\displaystyle =$ $\displaystyle \int \frac{4\sec^{2}{\theta}}{(4sec{\theta})^4}d\theta$  
  $\displaystyle =$ $\displaystyle \int \frac{d\theta}{4^3 \sec^{2}{\theta}} = \frac{1}{64} \int \cos^{2}{\theta}d\theta$  
  $\displaystyle =$ $\displaystyle \frac{1}{128}\int (1 + \cos{2\theta})d\theta = \frac{1}{128}(\theta + \frac{\sin{2\theta}}{2}) = \frac{1}{128}(\theta + \sin{\theta}\cos{\theta})$  
  $\displaystyle =$ $\displaystyle \frac{1}{128}(\tan^{-1}(\frac{x}{4}) + \frac{4x}{x^2 + 16})$  

(f) Let $x^3 = t$. Then $3x^2dx = dt$. Then

$\displaystyle \int{\frac{x^5}{x^9 - 1}} dx$ $\displaystyle =$ $\displaystyle \int{\frac{x^3 x^2dx}{(x^3)^3 - 1}}$  
  $\displaystyle =$ $\displaystyle \frac{1}{3}\int \frac{tdt}{t^3 - 1}$  

Note that

$\displaystyle \frac{t}{t^3 -1} = \frac{t}{(t-1)(t^2 + t+ 1)} = \frac{A}{t-1} + \frac{Bt + C}{t^2 + t + 1}$

Clear the denominator. Then

$\displaystyle t = A(t^2 + t+1) + (Bt+C)(t-1) = (A+B)t^2 +(A-B+C)t + (A -C)$

From this, we have the system of linear equations.

$\displaystyle \left\{\begin{array}{ll} A+B &= 0\\ A-B+C &= 1\\ A-C &=0 \end{array}\right.$

Solving this system by Cramer's rule. Then

$\displaystyle A = \frac{\left\vert \begin{array}{lll} 0 & 1 & 0\\ 1 & -1 & 1... ...1 & 1 & 0 \\ 1 & -1 & 1\\ 1 & 0 & -1 \end{array}\right\vert} = \frac{1}{3}$

Thus, $B = -\frac{1}{3}, C = \frac{1}{3}$ and
$\displaystyle \int{\frac{x^5}{x^9 - 1}} dx$ $\displaystyle =$ $\displaystyle \frac{1}{3}\int \frac{tdt}{t^3 - 1}$  
  $\displaystyle =$ $\displaystyle \frac{1}{9}\int{(\frac{1}{t-1} + \frac{-t + 1}{t^2 + t + 1})}dt$  
  $\displaystyle =$ $\displaystyle \frac{1}{9}\log\vert t-1\vert + \frac{1}{9}\int{\frac{-t + 1}{t^2 + t + 1}}dt$  

Now we write $\displaystyle{\frac{-t + 1}{t^2 + t + 1}}$ as follows:
$\displaystyle \int{\frac{-t + 1}{t^2 + t + 1}}dt$ $\displaystyle =$ $\displaystyle \int -\frac{1}{2}\int{\frac{2t - 2}{t^2 + t+1}}dt$  
  $\displaystyle =$ $\displaystyle -\frac{1}{2}\int{\frac{2t + 1 -3 }{t^2 + t+1}}dt = -\frac{1}{2}[\... ...ac{2t + 1}{t^2 + t+1}}dt + \int\frac{-3 }{(t + \frac{1}{2})^2 + \frac{3}{4}}dt]$  
  $\displaystyle =$ $\displaystyle -\frac{1}{2}[\log\vert t^2 + t+1\vert -3 \frac{2}{\sqrt{3}}\tan^{-1}{(\frac{2(t+\frac{1}{2})}{\sqrt{3}})}]$  

Thus,


$\displaystyle \int{\frac{x^5}{x^9 - 1}} dx$ $\displaystyle =$ $\displaystyle \frac{1}{3}\int \frac{tdt}{t^3 - 1}$  
  $\displaystyle =$ $\displaystyle \frac{1}{9}\log\vert t-1\vert + \frac{1}{9}(-\frac{1}{2})[\log\ve... ...t+1\vert -3 \frac{2}{\sqrt{3}}\tan^{-1}{(\frac{2(t+\frac{1}{2})}{\sqrt{3}})}]+c$  
  $\displaystyle =$ $\displaystyle \frac{1}{9}\log\vert x^3 -1\vert - \frac{1}{18}[\log\vert x^6 + x^3 + 1\vert + 2\sqrt{3} \tan^{-1}{(\frac{2x^3+1}{\sqrt{3}})}]+c$  

(g) Let $x^4 = t$. Then $4x^3dx = dt$．Thus,

$\displaystyle \int{\frac{1}{x(x^4 + 1)}} dx$ $\displaystyle =$ $\displaystyle \frac{1}{4}\int{\frac{4x^3 dx}{x^4(x^4 + 1)}}$  
  $\displaystyle =$ $\displaystyle \frac{1}{4}\int\frac{dt}{t(t+1)}$  

Now

$\displaystyle \frac{1}{t(t+1)} = \frac{A}{t} + \frac{B}{t + 1}$

Clear the denominator. Then we have

$\displaystyle 1 = A(t+1) + Bt = (A+B)t + A$

From this, we have the following system:

$\displaystyle \left\{\begin{array}{ll} A+B &= 0\\ A &= 1 \end{array}\right.$

Solve this system. Then we have $A = 1, B= -1$.
$\displaystyle \int{\frac{1}{x(x^4 + 1)}} dx$ $\displaystyle =$ $\displaystyle \frac{1}{4}\int\frac{dt}{t(t+1)} = \int (\frac{1}{t} + \frac{-1}{t + 1})dt$  
  $\displaystyle =$ $\displaystyle \frac{1}{4}[\log\vert t\vert - \log\vert t+1\vert] + c$  
  $\displaystyle =$ $\displaystyle \frac{1}{4}[\log\vert x^4\vert - \log\vert x^4+1\vert] + c$  

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Next: Integration of trigonometric functions Up: INTEGRATION Previous: Iintegration of rational functions