Question

∫arctanx（1+根号x）dx？

Answer 1

如图所示：

这题比较复杂：

如果你是说这个的话，比较简单：

Answer 2

I = ∫arctanx(1+√x)dx = ∫arctanxdx +∫√xarctanxdx
后者 令 u = √x，
I1 = ∫2u^2arctan(u^2)du = (2/3)∫arctan(u^2)d(u^3)
= (2/3)u^3arctan(u^2) - (2/3)∫2u^4/(1+u^4)du
= (2/3)u^3arctan(u^2) - (4/3)∫[1-1/(1+u^4)]du
= (2/3)u^3arctan(u^2) - 4u/3 +(4/3)∫[1/(1+u^4)]du
= (2/3)u^3arctan(u^2) - 4u/3 +(4/3)∫[1/(1+u^4)]du
∫[1/(1+u^4)]du = [1/(4√2)]ln|(u^2-√2u+1)/(u^2+√2u+1)| 代入得
I1 = (2/3)u^3arctan(u^2) - 4u/3 +[1/(3√2)]ln|(u^2-√2u+1)/(u^2+√2u+1)|
= (2/3)x^(3/2)arctanx - 4√x/3 + [1/(3√2)]ln|[x-√(2x)+1]/[x+√(2x)+1]|
I = xarctanx - ∫xdx/(1+x^2) + I1

= xarctanx - (1/2)∫d(1+x^2)/(1+x^2) + I1
= xarctanx - (1/2)ln(1+x^2) + (2/3)x^(3/2)arctanx - 4√x/3
+ [1/(3√2)]ln|[x-√(2x)+1]/[x+√(2x)+1]| + C

Answer 3

∫arctan(1+√x)dx