Question

Resolver la integral indefinida: ∫[e×/(e×-1)(e×+4)] dx
Nota: Es por fracciones parciales.

Answer 1

Tienes la integral,

$\displaystyle\int{\frac{e^{x}}{(e^{x}-1)(e^{x}+4)}}dx\\\\Sea,s=e^{x},ds=e^{x}dx,entonces,dx=\frac{ds}{e^{x}},\\\\\int{\frac{e^{x}}{(s-1)(s+4)}}\left(\frac{ds}{e^{x}}\right)=\int{\frac{ds}{(s-1)(s+4)}}$

y listo, ahora si puedes usar fracciones parciales, tienes factores lineales, que no se repiten y reales, entonces,

$\dfrac{1}{(s-1)(s+4)}=\dfrac{A}{s-1}+\dfrac{B}{s+4}=\dfrac{A(s+4)+B(s-1)}{(s-1)(s+4)}\\...=\dfrac{As+4A+Bs-B}{(s-1)(s+4)}=\dfrac{(A+B)s+(4A-B)}{(s-1)(s+4)}\\\\\dfrac{1}{(s-1)(s+4)}=\dfrac{(A+B)s+(4A-B)}{(s-1)(s+4)}\\1=(A+B)s+(4A-B)\\\\ \displaystyle\left \{ {{A+B=0} \atop {4A-B=1}} \right. =\left \{ {{A=-B} \atop {4A-B=1}} \right.=\left \{ {{A=\frac{1}{5}} \atop {B=-\frac{1}{5}}} \right.$

entonces la fracciones parciales te quedarían

$\displaystyle\frac{1}{5}\int{\left(\frac{1}{s-1}-\frac{1}{s+4}\right)ds}=\frac{1}{5}\big(\ln|s-1|-\ln|s+4|\big)=\frac{1}{5}\ln\left|\frac{s-1}{s+4}\right|\\\\...=\frac{1}{5}\ln\left|\frac{e^{x}-1}{e^{x}+4}\right|+C$

y eos sería todo¡