求不定积分∫xdx/(4x^2+4x+5)
2个回答

令t=2x+1,x=(t-1)/2 dx=dt/2

∫xdx/(4x^2+4x+5)

=∫(t-1)/2 *dt/2 /(t^2+4)

=1/8*∫(2t-2)/(t^2+4) *dt

=1/8 *∫1/(t^2+4)*d(t^2+4)-1/4*∫dt/(t^2+4)

=1/8*ln(t^2+4)-1/8*∫d(t/2)/((t/2)^2+1)

=1/8*ln(t^2+4)-1/8arctg(t/2)+c

1/8ln(4x^2+4x+5)-1/8arctg(x+1/2)+c