Đặt `t=sqrt(e^x+1)=>dt=(e^x)/(2sqrt(e^x+1))dx`

`=>` `dx=(2sqrt(e^x+1))/(e^x)dt=2sqrt(e^x+1)*e^(-x)dt`

`=>` `dx=2t*e^(-x)dt`

Đồng thời ta có: `t^2=e^x+1=>e^x=t^2-1`

Ta có: `I=int_(ln3)^(ln15) 1/(e^(-x)(sqrt(e^x+1)+e^x-1))dx`

`=int_(2)^4 1/(e^(-x)(t+t^2-1-1))*2t*e^(-x)dt`

`=int_2^4 (2t)/(t^2+t-2)dt`

`=int_2^4 (2t)/((t-1)(t+2))dt`

`=int_2^4 ((2/3)/(t-1)+(4/3)/(t+2))dt`

`=2/3int_2^4 1/(t-1)dt+4/3int_2^4 1/(t+2)dt`

`=2/3*(ln |t-1|)|_2^4``+4/3*(ln |t+2|)|_2^4`

`=2/3ln3+4/3*(ln6-ln4)`

`=2/3ln3+4/3*(ln3+ln2-2ln2)`

`=2/3ln3+4/3*(ln3-ln2)`

`=2ln3-4/3ln2`

`=>` `a=2;b=3;c=4;d=3;k=2`

`=>` `a+c+d=9`

`=>` $\fbox{CHỌN B}$

Lưu ý: Đề bài có thể đã nhầm vì `b=3 > k=2`