Question

急需！！！！！ 求不定积分 谢谢各位高手

∫（x^2-3)(x+1)/x^2 dx
∫ x-9/根号X +3 dx

Answer 1

(1)∫(x²-3)(x+1)/x² dx
=∫(x³+x²-3x-3)/x² dx
=∫xdx+∫dx-3∫1/x dx-3∫1/x² dx
=x²/2+x-3ln|x|+3/x+C

(2)∫(x-9)/√(x+3) dx,u=x+3,du=dx
=∫(u-3-9)/√u du
=∫(u-12)/√u du,t=√u,u=t²,du=2tdt
=∫(t²-12)/t*2t dt
=2∫(t²-12)dt
=2∫t²dt-24∫dt
=2*t³/3-24t+C
=2/3*u^(3/2)-24√u+C
=2/3(x+3)^(3/2)-24√(x+3)+C

Answer 2

∫（x^2-3)(x+1)/x^2 dx
=∫(x^3+x^2-3x-3)/x^2 dx
=∫(x+1-3/x-3/x^2)dx
=x^2/2+x-3ln|x|+3/x+C

∫ x-9/根号X +3 dx
=x^2/2-18根号X+3x+C

Answer 3

(1)x^2/2+x-3ln|x|+3/x+C
(2)x^2/2-18根号X+3x+C