Question

求不定积分∫(1/x^2+2x+5)dx

有搜到答案是这样的，但是不懂怎么从倒数第二步到最后一步
∫1/(x^2+2x+5)dx
=∫1/[(x+1)^2+4]dx
=∫1/[(x+1)^2+2^2]d(x+1)
=(1/2)arctan[(x+1)/2]+C

Answer 1

∫1/(x^2+2x+5)dx
=∫1/[(x+1)^2+4]dx
分子分母同除以4
=∫(1/4)/[(x/2+1/2)^2+1]dx
=(1/4)*2∫1/[(x/2+1/2)^2+1]d(x/2+1/2)
=1/2∫1/[(x/2+1/2)^2+1]d(x/2+1/2)
=1/2arctan[(x+1)/2]+C
明白?可继续问.
附:arctanx'=1/(1+x^2)

Answer 2

第二步就配平方，第三步换元，
∫ dx/(a^2 + x^2) = (1/a)arctan(x/a) + C

Answer 3

=∫1/[(x+1)^2+2^2]d(x+1)
=∫1/2^2{(x+1)/2]^2+1}d(x+1) 在分母把2^2提出来
=1/4∫1/{(x+1)/2]^2+1}d(x+1)
=1/2∫1/{(x+1)/2]^2+1}d(x+1)/2
=(1/2)arctan[(x+1)/2]+C ( 有公式 (arctanx)'=1/(x^2+1) )

Answer 4

微分里面需要凑成d(x+1)/2

Answer 5

∫1/(x²+2x+5)dx
=∫1/[(x+1)²+4]dx
=∫1/[(x+1)²+2²]d(x+1)
=∫(1/4)/([(x+1)/2]²+1)

=(1/2)∫d[(x+1)/2]/([(x+1)/2]²+1)
=(1/2)arctan[(x+1)/2]+C