如何求x*(1.08^-x)的不定积分

数学符号在电脑上写太麻烦 我直接写在纸上上传了  你看看吧

S x[1.08(^-x)]dx

= S xe^[-xln(1.08)]dx

= -[ln(1.08)]^(-1)S xd{e^[-xln(1.08)]}

= -x[ln(1.08)]^(-1)e^[-xln(1.08)] + [ln(1.08)]^(-1)S e^[-xln(1.08)]dx

= -x[ln(1.08)]^(-1)e^[-xln(1.08)] - [ln(1.08)]^(-2)S d{e^[-xln(1.08)]}

= -x[ln(1.08)]^(-1)e^[-xln(1.08)] - [ln(1.08)]^(-2)e^[-xln(1.08)] + C.

C = const.