[tex] \left \begin{array}{rcl} t=1 & & 2B \ = \ 4 \ e^{-\tan \left ( \frac{1}{2} \right ) } \\t=0 & & -A+B-D \ = 1 \\ t=-1 & & -4A+2B-4C+4D \ = 0 \\ t=2 & & 5A+5B+2C+D \ =\ 9 \ e^{-\tan (1)}[/tex]
Despejando y demás (comprobadlo por dió!), me da:
[tex]\begin{array}{rcl} B=2 \ e^{-\tan \left ( \frac{1}{2} \right ) } \\ \\ A=\frac{42 \ e^{-\tan \left ( \frac{1}{2} \right ) } - 9 e^{-\tan (1) } }{4} \\ \\D= \frac{-34 \ e^{-\tan \left ( \frac{1}{2} \right ) }+ 9 \ e^{-\tan (1) } }{4} \\ \\ C=\frac{9 \ e^{-\tan (1) } -36 \ e^{-\tan \left ( \frac{1}{2} \right ) } }{2}[/tex]
[tex] \frac{1}{2} \int \frac{ ( \ 18 e^{-\tan (1) }-72 e^{-\tan \left ( \frac{1}{2} \right ) } ) \ t- \ 34 e^{-\tan \left ( \frac{1}{2} \right ) } \ + \ 9 \ e^{-\tan \left ( \frac{1}{2} \right ) } }{1+t^2} [/tex]
[tex]=-e^{-\tan \left ( \frac{1}{2} \right ) } \int \frac{36t+17}{1+t^2} + \frac{9}{2} e^{-\tan (1) } \int \frac{1+2t}{1+t^2}[/tex]
Hallando las integrales, deshaciendo el cambio y agrupando:
[tex]-18 e^{-\tan \left ( \frac{1}{2} \right ) } \ln |1+t^2|-17 e^{-\tan \left ( \frac{1}{2} \right ) } \arctan |t| +\frac{9}{2} e^{-\tan (1) } \arctan |t|+\frac{9}{2} \ln |1+t^2|[/tex]
[tex] \boxed{ \left ( -18 \ e^{-\tan \left ( \frac{1}{2} \right ) } \ + \ \frac{9}{2} e^{-\tan (1) } \right ) \ln|1+2 \ \arctan(e^x)| \ + \ \arctan | 1+ \ 2 \ \arctan (e^x) | \left ( \frac{9}{2} e^{-\tan (1)} -17 e^{-\tan \left ( \frac{1}{2} \right ) } \right ) \ + \ K}[/tex]
Sale un chorizaco. No tengo ni idea de si esta bien o mal, pero ahi queda
