Question

求不定积分∫xdx/(4x^2+4x+5)

谢谢
1楼应该是对的
2楼的思路不错但第一行就错...下面更正
本题的关键：4x²+4x+5=(2x+1)²+4
还有对4x²+4x+5求导得8x+4,所以才会想到添项+4-4
∫xdx/(4x²+4x+5)
=1/8∫(8x+4-4)dx/(4x²+4x+5)
=1/8∫(8x+4)dx/(4x²+4x+5)-1/8∫4 dx/(4x²+4x+5)
=1/8∫d(4x²+4x+5)/(4x²+4x+5)-1/2∫dx/[(2x+1)²+4]
=1/8ln|4x²+4x+5|-1/2∫dx/[(2x+1)²+4]
=1/8ln|4x²+4x+5|-1/8∫d(x+1/2)/[(x+1/2)²+1]
=1/8ln|4x²+4x+5|-1/8arctan(x+1/2)+C

Answer 1

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

Answer 2

4x²+4x+5=4x²+4x+1+4=(2x+1)²+1
∫xdx/(4x²+4x+5)
=1/8∫(8x+4-4)dx/(4x²+4x+5)
=1/8∫(8x+4)dx/(4x²+4x+5)-1/8∫4 dx/(4x²+4x+5)
=1/8∫d(4x²+4x+5)/(4x²+4x+5)-1/2∫dx/[(2x+1)²+1]
=1/8ln|4x²+4x+5|-1/2∫dx/[(2x+1)²+1]
=1/8ln|4x²+4x+5|-1/4∫d(2x+1)/[(2x+1)²+1]
=1/8ln|4x²+4x+5|-1/4arctan(2x+1)+C