Question

求下列不定积分？

求下列不定积分，谢谢🙏

Answer 1

网上真多乱七八糟的答案……现在上图……如图所示

Answer 2

1) ∫(4/x^(1/2))*sin(x^(1/2))dx = 8∫sin(x^(1/2))d x^(1/2 )=-8 ∫dcos(x^(1/2))=-8cos(x^(1/2)) + C
2) 令t=(x+2)^(1/2), 那么x=t^2-2, dx=2tdt,
∫x (x+2)^(1/2)dx= ∫(t^2-2)t*2tdt= 2 ∫(t^4-2t^2)dt= 2 ∫d[(t^5)/5-(2t^3)/3)]

=2t^5/5-4t^3/3 + C= 2 (x+2)^(5/2)/5-4 (x+2)^(3/2)/3 + C
3) ∫x (x^2+1)^(1/2)dx=1/2 ∫ (x^2+1)^(1/2)d(x^2+1)= 1/3 ∫d(x^2+1)^(3/2) = (x^2+1)^(3/2) /3 + C
4) 令t=(x+6)^(1/2), x=t^2-6, dx=2tdt
∫1/[1+(x+6)^(1/2) dx = ∫[2t/(1+t)]dt=2∫[1-1/(1+t)]dt=2t - 2ln(1+t) + C
=2 (x+6)^(1/2) - 2ln[1 +(x+6)^(1/2)] +C