第一步:展开完全平方项。
(x^2+1)^2=x^4+2x^2+1
第二步:将x乘以上式,并将2提到不定积分外。
被积函数为(x^5+2x^3+x)/(x+1)
第三步:将上式展开。
得x^4-x^3+3x^2-3x+4x/(x+1)
第四步:分别积分,然后求和。
详见附图:
1/[(x+1)^2.(x^2+1)]≡ A/(x+1) +B/(x+1)^2 +(Cx+D)/(x^2+1)
=>
1≡A(x+1)(x^2+1) +B(x^2+1) +(Cx+D)(x+1)^2
x=-1, => B=1/2
x=i
(Ci +D)(i+1)^2 = 1
(Ci +D)(2i) = 1
-2C +2Di =1
=> C=-1/2 and D=0
coef. of constant
A+B+D=1
A+1/2 +0=1
A=1/2
1/[(x+1)^2.(x^2+1)]
≡ A/(x+1) +B/(x+1)^2 +(Cx+D)/(x^2+1)
≡ (1/2)[1/(x+1)+ 1/(x+1)^2 -x/(x^2+1) ]
∫2x/[(x+1)(x^2+1)^2] dx
=-∫[1/(x+1)] d[1/(x^2+1)]
=-1/[(x+1)(x^2+1)] -∫dx/[(x+1)^2.(x^2+1)]
=-1/[(x+1)(x^2+1)] -(1/2)∫[1/(x+1)+ 1/(x+1)^2 -x/(x^2+1) ] dx
=-1/[(x+1)(x^2+1)] -(1/2)[ ln|x+1| - 1/(x+1) -(1/2)ln|x^2+1| ] + C
(x²+1)² 乘到分母上 ? 分子上 ?
不定积分结果不唯一求导验证应该能够提高凑微分的计算能力先写别问唉。。类似