Question

考研定积分问题求大佬

如图

Answer 1

需要对微分真正的理解，微分一个重要的性质，就是微分形式的不变性。

Answer 2

麻烦的地方在分母那多出了一个x，这种情况一般用倒代换

即令t=1/x。这样就可以把分母的x去掉了后面就好做了。具体如下

Answer 3

Answer 4

解：分享一种解法。∵3x^4+x²+1=3[(x²+1/6)²+11/36]，设x²+1/6=[(√11)/6]tanθ，
∴x²=[(√11)/6]tanθ-1/6=-√(1/3)secθcos(θ+α)【α=arctan(√11)】。
∴∫dx/[x√(3x^4+x²+1)]=(-1/2)∫sec(θ+α)dθ=(-1/2)ln丨sec(θ+α)+tan(θ+α)丨+C。
∴原式=(1/2)ln[11/(3+2√5)]。

供参考。

Answer 5

分子分母同乘以 2x，
I = ∫<√2/2, 1>dx^2/[2x^2√(1+x^2+3x^4)], 令 u = x^2
= ∫<1/2, 1>du/[2u√(1+u+3u^2
= ∫<1/2, 1>du/[2u√[11/12 + 3(u+1/6)^2],
= ∫<1/2, 1>du/[u√[11/3 + 12(u+1/6)^2],
= √(3/11)∫<1/2, 1>du/[u√[1 + (36/11)(u+1/6)^2],
令 (6/√11)(u+1/6) = tanv, 则 u = (√11/6)tanv - 1/6, du = (√11/6)(secv)^2dv
I = (√3/6)∫<arctan(4/√11), arctan(7/√11)>secvdv/[(√11/6)tanv-1/6]
= √3∫<arctan(4/√11), arctan(7/√11)>secvdv/(√11tanv-1)

= √3∫<arctan(4/√11), arctan(7/√11)>dv/(√11sinv-cosv)
= (1/2)∫<arctan(4/√11), arctan(7/√11)>dv/sin(v-t), 其中 tant = 1/√11
= (1/2)∫<arctan(4/√11), arctan(7/√11)>csc(v-t)d(v-t)
= (1/2)[ln{csc(v-t) - cot(v-t)}]<arctan(4/√11), arctan(7/√11)>
= (1/2)[ln{1/(sinvcost-cosvsint) - (1+tanvtant)/(tanv-tant)}]<arctan(4/√11), arctan(7/√11)>
= (1/2)ln{1/[(7/√60)(√11/√12)-(√11/√60)(1/√12)] - [1+(7/√11)(1/√11)]/(7/√11+1/√11)}
-(1/2)ln{1/[(4/√27)(√11/√12)-(√11/√27)(1/√12)]-[1+(4/√11)(1/√11)]/(4/√11+1/√11)}
= (1/2){ln(8√5/√11)] - ln[7/(30√11)]}
= (1/2)[ln8 + (1/2)ln5 - (1/2)ln11 - ln7 + ln30 + (1/2)ln11]
= (1/2)[3ln2 + (1/2)ln5 - ln7 + ln5 + ln2 + ln3 ]
= 2ln2 +(3/4)ln5 + (1/2)ln3 - (1/2)ln7