Question

求不定积分(X-2)/(X^2+2X+3)^2 哪位亲帮我回答一下，要写出详细的步骤哦。谢谢了

Answer 1

点击放大，荧屏放大再放大：

Answer 2

=∫(x+1)dx/(x^2+2x+3)^2 -3∫dx/(x^2+2x+3)^2
=(-1/[2(x^2+2x+3)]-3∫d(x+1)/[(x+1)^2+2]^2
∫d(x+1)/[(x+1)^2+2]^2
√2tgU=x+1 U=arctg[(x+1)/√2],
d(x+1)=√2(secU)^2dU
∫√2(secU)^2dU/4secU^4
=(√2/4)∫(cosU^2)dU
=(√2/8)∫(1+cos2U)dU
=√2/8U+√2/16sin2U
原式=-1/[2(x^2+2x+3)]-(3√2/8)arctg[(x+1)/√2]-(3√2/16)sin[2arctg[(x+1)/√2]]
tgU=(x+1)/√2, secU^2=[(x+1)^2/2]+1=(x^2+2x+3)/2
cosU^2=2/(x^2+2x+3) sinU^2=(x^2+2x+1)/(x^2+2x+3)
sin2U=2sinUcosU=2√2(x+1)/(x^2+2x+3)
代入
原式=-1/[2(x^2+2x+3)]-(3√2/8)arctg[(x+1)/√2]-(12/16)(x+1)/(x^2+2x+3)
=-5/[4(x^2+2x+3)]-(3x)/[4(x^2+2x+30]-(3√2/8)arctg[(x+1)/√2]