Integration of rational functions

(C.1).
Partial fractions

$ \displaystyle\frac{1}{x^3 + 1}
= \frac{1}{3}\left[\frac{1}{x + 1} - \frac{x - 2}{x^2 - x+1}\right]$ ; $ \displaystyle\frac{1}{x^3 - 1}
= \frac{1}{3}\left[\frac{1}{x - 1} - \frac{x + 2}{x^2 + x+1}\right]$ .
$ \displaystyle\frac{1}{x^4 - 1}
= \frac{1}{2}\left[\frac{1}{x^2-1} - \frac{1}{x^2+1}\right]$ ; $ \displaystyle\frac{1}{x^4 + 1}
= \frac{\sqrt{2}}{4}\left[
\frac{x+\sqrt{2}}{x^2+\sqrt{2}x+1} - \frac{x-\sqrt{2}}{x^2-\sqrt{2}x+1}
\right]^\star$ .


$ \displaystyle\left[
^\star\mbox{ Use }\:
x^4 + 1 = (x^2+1)^2 - (\sqrt{2} x)^2
= (x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)
\:\right]$

  (a) $\displaystyle \int \frac{1}{x^3 + 8} dx = \frac{1}{4}\int \frac{1}{t^3 + 1} dx \hspace{0.8in}\fbox{(A.2) $x = 2t$,\: $\frac{dx}{dt} = 2$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \frac{1}{12}\int \left(\frac{1}{t+1} - \frac{t-2}{t^2-t+1}\right) dt \hspace{0.8in}\fbox{(C.1)}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \frac{1}{12}\ln\vert t+1\vert -\frac{1}{12}\left...
...{3}{2}\int\frac{1}{t^2-t+1} dt\right] \hspace{0.8in}\fbox{(C.2)}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \frac{1}{12}\ln\vert t+1\vert -\frac{1}{24}\ln\v...
...\left(\frac{2t-1}{\sqrt{3}}\right) + C \hspace{0.8in}\fbox{(C.3)}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \frac{1}{12}\ln\left\vert\frac{x}{2}+1\right\ver...
...1\right\vert +\frac{\sqrt{3}}{12}\tan^{-1}\left(\frac{x-1}{\sqrt{3}}\right) + C$    

(C.2).
$ \displaystyle\int \frac{x+a}{x^2+bx+c} dx
= {\textstyle\left(\frac{1}{2}\righ...
...c\vert^{\:*}
+ {\textstyle\left(a-\frac{b}{2}\right)}\int\frac{1}{x^2+bx+c} dx$


$ \left[
^{*}\mbox{ Use }\:
\frac{x+a}{x^2+bx+c} =
{\textstyle\left(\frac{1}{2}\...
...{ and (A.1) with }\:
t = x^2+bx+c,\:
{\textstyle\frac{dt}{dx} = 2x+b}\:
\right]$


(C.3).
$ \displaystyle\int \frac{1}{x^2+bx+c} dx
= \begin{cases}
\frac{2}{\vert b^2-4c...
... b^2-4c\vert^\frac{1}{2}}\right)^{**}
& \mbox{ if $b^2 - 4c > 0$. }
\end{cases}$


$ \left[
^{**}\mbox{ Use }\:
\frac{1}{x^2+bx+c} = \frac{4}{(2x+b)^2-(b^2-4c)}
\:...
...:
{\textstyle\frac{dx}{dt} = \frac{\vert b^2-4c\vert^\frac{1}{2}}{2}}\:
\right]$

  (a) $\displaystyle \int \frac{1}{x^2 + x + 1} dx = \frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{2x+1}{\sqrt{3}}\right) + C \hspace{0.8in}\fbox{(C.3)}\hspace{5.0in}$    

  (b) $\displaystyle \int \frac{x}{x^4+x^2+1} dx = \int \frac{x}{(x^2+\frac{1}{2})^2+...
....8in}\fbox{(A.1) $t = x^2 + \frac{1}{2}$,\: $\frac{dt}{dx} = 2x$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \frac{1}{\sqrt{3}} \tan^{-1} \left(\frac{2t}{\sqrt{3}}\right) + C \hspace{0.8in}\fbox{(C.3)}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \frac{1}{\sqrt{3}} \tan^{-1} \left(\frac{2x^2+1}{\sqrt{3}}\right) + C$    


(C.4).
$ \displaystyle\int \frac{1}{(1 + x^2)^{n+1}} dx
= \frac{x}{2n(1 + x^2)^n}
+ \frac{2n-1}{2n} \int \frac{1}{(1 + x^2)^n} dx^{***}$

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


$ \displaystyle\left[
^{***}\mbox{ Use }\:
\frac{d}{dx}\left(\frac{x}{(1+x^2)^n}\right)
= -\frac{2n-1}{(1+x^2)^n} + \frac{2n}{(1+x^2)^{n+1}}\:
\right]$

  (a) $\displaystyle \int \sec^5 x dx = \int \frac{1}{\cos^5 x} dx = \int \frac{\cos...
...space{0.4in}\fbox{(A.1) $t = \sin x$,\: $\frac{dt}{dx} = \cos x$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \frac{t}{4(1 - t^2)^2} + \frac{3}{4}\int \frac{1...
... \frac{1}{(1 - t^2)} dt \right] \hspace{0.2in}\fbox{(C.4) twice}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \frac{t}{4(1 - t^2)^2} + \frac{3t}{8(1 - t^2)} +...
...in x}{4\cos^4 x} + \frac{3\sin x}{8\cos^2 x} + \frac{3}{8}\tanh^{-1} \sin x + C$    

  (b) $\displaystyle \int \frac{1}{(x^2 + 4)^3} dx = \frac{1}{32} \int \frac{1}{(t^2 ...
...3} dt \hspace{0.8in}\fbox{(A.2) $x = 2t$,\: $\frac{dx}{dt} = 2$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.2in} = \frac{1}{32}\left[ \frac{t}{4(t^2 + 1)^2} + \fra...
...{1}{t^2 + 1} dt \right]\right] \hspace{0.05in}\fbox{(C.4) twice}\hspace{5.0in}$    
  $\displaystyle \hspace{0.2in} = \frac{1}{32}\left[ \frac{t}{4(t^2 + 1)^2} + \fra...
...6(x^2 + 4)^2} + \frac{3x}{128(x^2 + 4)} + \frac{3}{256}\tan^{-1}\frac{x}{2} + C$    



Department of Mathematics
Last modified: 2005-10-05