Question

求不定积分∫x/√(1+x+x^2)dx

Answer 1

直接用公式法，简单快捷

原式=√(1+x+x^2) -(1/2)ln|2√(1+x+x^2) + 2x+1| + C

Answer 2

x^2+x+1 = (x+1/2)^2+ 3/4

let

x+1/2 = (√3/2)tanu

dx =(√3/2)(secu)^2 du

∫x/√(1+x+x^2)dx

=(1/2)∫(2x+1)/√(1+x+x^2)dx -(1/2)∫dx/√(1+x+x^2)

=√(1+x+x^2) -(1/2)∫dx/√(1+x+x^2)

=√(1+x+x^2) -(1/2)∫ secu du

=√(1+x+x^2) -(1/2)ln|secu + tanu| + C'

=√(1+x+x^2) -(1/2)ln|(2/√3)√(1+x+x^2) + (2x+1)/√3 | + C'

=√(1+x+x^2) -(1/2)ln|2√(1+x+x^2) + (2x+1)| + C

Answer 3

😁

追答