数学才子昂。求不定积分。跪求详细步骤！！！

Question

1.∫[√x]/[√x-(√x)^3]dx
2.∫x^2ln（x+1）dx
3.∫[e^（ax）]cosbxdx

Answer 1

1、∫ √x / [√x - x^(3/2)] dx
= ∫ 1/(1-x) dx
= -∫ d(1-x)/(1-x)
= -ln|1-x| + C
2、∫ x²ln(x+1) dx
= ∫ ln(x+1) d(x³/3)
= (1/3)x³ln|x+1| - (1/3)∫ x³ * 1/(x+1) dx
= (1/3)x³ln|x+1| - (1/3)∫ [x²-x-1/(x+1)+1] dx
= (1/3)x³ln|x+1| - (1/3)[x³/3-x²/2-ln|x+1|+x] + C
= (1/3)(x³+1)ln|x+1| - (x/18)(2x²-3x+6) + C
3、∫ e^(ax) * cos(bx) dx
= (1/b)∫ e^(ax) d(sinbx)
= (1/b)e^(ax)sinbx - (1/b)∫ sinbx * ae^(ax) dx
= (1/b)e^(ax)sinbx + (a/b)(1/b)∫ e^(ax) d(cosbx)
= (1/b)e^(ax)sinbx + (a/b²)e^(ax)cosbx - (a/b²)∫ cosbx * ae^(ax) dx
= (1/b)e^(ax)sinbx + (a/b²)e^(ax)cosbx - (a²/b²)∫ e^(ax) * cos(bx) dx
(1+a²/b²)∫ e^(ax) * cos(bx) dx = (1/b)e^(ax)sinbx + (a/b²)e^(ax)cosbx
∫ e^(ax) * cos(bx) dx = [(1/b)e^(ax)sinbx + (a/b²)e^(ax)cosbx] / (1+a²/b²) + C
= e^(ax) * [a*cos(bx) + b*sin(bx)] / (a²+b²) + C