對(duì)lnx的積分,唯有分部積分法
∫ lnx dx = x * lnx - ∫ x d(lnx)
= xlnx - ∫ x * 1/x dx
= xlnx - x + C
對(duì)a^x的積分,用∫ e^(kv) dv = (1/k)e^(kv)
則∫ a^x dx
= ∫ e^[ln(a^x)] dx,有ƒ(x) = e^[lnƒ(x)]
= ∫ e^(x * lna) * 1/lna * (lna dx)
= (1/lna)∫ e^(x * lna) d(x * lna)
= (1/lna) * e^(x * lna) + C
= (a^x)/lna + C
或令y = a^x
lny = ln(a^x) = x * lna,兩邊求導(dǎo)
1/y * dy/dx = 1 * lna
dy/dx = y * lna = a^x * lna,兩邊取積分
y + C' = ∫ a^x * lna dx
即lna * ∫ a^x dx = a^x + C'
==> ∫ a^x dx = (a^x)/lna + C,其中(C = C'/lna)