不定积分公式推导

答：x^(t+1)/(t+1)
+
c
假设x是变量，n∈r.
这要由导数开始推导：
考虑函数y
=
xⁿ
则y'
=
nxⁿ⁻¹
因为
(xⁿ)'_x
=
lim(δx->0)
[
(x+δx)ⁿ
-
xⁿ
]/δx，分子运用二项式定理展开
=
lim(δx->0)
[
(xⁿ+nxⁿ⁻¹δx+o(δx))
-
xⁿ
]/δx
=
lim(δx->0)
(nxⁿ⁻¹δx+o(δx))/δx
=
lim(δx->0)
[
nxⁿ⁻¹+o(δx)
]，o(δx)为比δx更高阶的项
=
nxⁿ⁻¹
把n替换为n+1
即(xⁿ⁺¹)'_x
=
(n+1)xⁿ
即[xⁿ⁺¹/(n+1)]'_x
=
xⁿ
所以两边取不定积分，有
∫
xⁿ
dx
=
xⁿ⁺¹/(n+1)
+
c，c为任意常数项

左边=∫dx/cosx=∫cosxdx/(cosx)^2
=∫d(sinx)/[1-(sinx)^2]
令t=sinx,
=∫dt/(1-t^2)
=(1/2)∫dt/(1+t)+(1/2)∫dt/(1-t)
=(1/2)∫d(1+t)/(1+t)-(1/2)∫d(1-t)/(1-t)
=(1/2)ln|1+t|-(1/2)ln|1-t|+C
=(1/2)ln|(1+t)/(1-t)|+C
=(1/2)ln|(1+sinx)/(1-sinx)|+C
//在对数中分子分母同乘1+sinx,
=(1/2)ln|(1+sinx)^2/(cosx)^2|+C
=ln|(1+sinx)/cosx|+C
=ln|1/cosx+sinx/cosx|+C
=ln(secx+tanx|+C=右边，
∴等式成立。