求不定积分.二次根下.根号里面是,e的x次方加1.求详细解答.好人一...求不定积分.二次根下.根号里面是,e的x次方加1.求详细解答.好人一生平安!

Question

求不定积分.二次根下.根号里面是,e的x次方加1.求详细解答.好人一...
求不定积分.二次根下.根号里面是,e的x次方加1.求详细解答.好人一生平安!

Answer 1

∫ √(e^x+1)dx
令：√(e^x+1)=t
e^x+1=t^2
x=ln(t^2-1)
dx=(t^2-1)^(-1)*(2t)
原式=∫ 2t^2/(t^2-1)dt
=2∫ t^2/(t^2-1)dt
=2[∫ 1+(t^2-1)^(-1)]dt
=2t+2∫ 1/(t^2-1)dt
=2t+ln|(t-1)/(t+1)|+C
=2√(e^x+1)+ln|e^x/(e^x+2)|+C.
=2√(e^x+1)+x-ln(e^x+2)+C