Question

(1)x乘以(2x+3)的4次开根号的不定积分怎么求

Answer 1

1）∫√[x*(2*x+3)^4]dx

=∫(x^(1/2)*(2x+3)^2)*2*dx^(1/2) （dx化为 2*x^(1/2)dx^(1/2)）

=2∫(4x^3 +12*x^2 +9x)dx^(1/2)

=2∫4x^3 dx^(1/2)+2∫12*x^2 dx^(1/2)+2∫9x dx^(1/2)

=(8/7)*x^(7/2)+(24/5)*x^(5/2)+6*x^(3/2)