令t=√[e^(2x)+5],得x=1/2·ln(t²-5),dx=t/(t²-5) dt
∫√[e^(2x)+5] dx
=∫t·t/(t²-5) dt
=∫(t²-5+5)/(t²-5) dt
=∫[1+5/(t²-5)]dt
=∫dt+1/2·∫[1/(t-5)-1/(t+5)]dt
=t+1/2·(ln|t-5|-ln|t+5|)+C
=t+1/2ln|(t-5)/(t+5)|+C
=√[e^(2x+5)]+1/2 ln|{√[e^(2x)+5]-5}/{√[e^(2x)+5]+5}|+C