作业帮如何求下面三个式子的不定积分?一、1/((x^4)*(1+x^2))二、(x^2+2)/(1+x+x^2)^2三、((1+x)/(1-x))^0.5
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如何求下面三个式子的不定积分?一、1/((x^4)*(1+x^2))二、(x^2+2)/(1+x+x^2)^2三、((1+x)/(1-x))^0.5
如何求下面三个式子的不定积分?
一、1/((x^4)*(1+x^2))
二、(x^2+2)/(1+x+x^2)^2
三、((1+x)/(1-x))^0.5
如何求下面三个式子的不定积分?一、1/((x^4)*(1+x^2))二、(x^2+2)/(1+x+x^2)^2三、((1+x)/(1-x))^0.5
1
∫dx/(x^4 *(1+x^2))
=∫[(1+x^2)^2-2x^2-x^4]dx/[x^4*(1+x^2)]
=∫(1+x^2)dx/x^4-∫2dx/[x^2(1+x^2)]-∫dx/(1+x^2)
=(-1/3)x^(-3)+(-1)x^(-1)-2∫(1+x^2-x^2)dx/[x^2(1+x^2)]-∫dx/(1+x^2)
=-1/3)x^(-3)+x^(-1)-3arctanx+C
2
∫(x^2+2)dx/(1+x+x^2)^2
=∫(x^2+x+1-x-1)dx/(1+x+x^2)
=∫dx/(1+x+x^2)-∫(x+1)dx/(1+x+x^2)^2
=∫dx/[(x+1/2)^2+3/4]-(1/2)∫d(x^2+x+1)/(1+x+x^2)^2 -(1/2)∫dx/[(x+1/2)^2+3/4]
=(1/2)*(√3/2)arctan(√3x/2+√3/4)+(1/2)(1+x+x^2)^(-1)+C
3
∫(1+x)^(1/2)dx/(1-x)^(1/2)
设x=cosu, dx=-sinudu
=∫cos(u/2)(-sinu)du/sin(u/2)
=∫-2(cosu/2)^2du
=∫-(1+cosu)du
=-u-sinu+C
=-arccosx-√(1-x^2)+C
1. 1/((x^4)*(1+x^2)) = [(1+x^2)-x^2] / [ (x^4)*(1+x^2)]
= x^(-4) – x^(-2) – 1/(1+x^2)
I = (-1/3)x^(-3) + 1/x – arctanx +C
2. (x^2+2)/(1+x+x^2)^2 = 1/(1+x+x^2) + [–(1/2)(2x+1)+3/2]/ (...
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1. 1/((x^4)*(1+x^2)) = [(1+x^2)-x^2] / [ (x^4)*(1+x^2)]
= x^(-4) – x^(-2) – 1/(1+x^2)
I = (-1/3)x^(-3) + 1/x – arctanx +C
2. (x^2+2)/(1+x+x^2)^2 = 1/(1+x+x^2) + [–(1/2)(2x+1)+3/2]/ (1+x+x^2)^2
I = ∫1/(1+x+x^2)dx + (1/2)/ (1+x+x^2) + (3/2)∫1/(1+x+x^2)^2 dx
1+x+x^2 = (x+1/2)^2+3/4,用换元法，令x+1/2 =√3 tant /2
……
= (1+x)/(1+x+x^2) + 4/√3 arctan[(2x+1)/√3] + C
3. √【(1+x)/(1-x)】= (1+x)/√(1-x^2)
I = ∫[1/√(1-x^2) + x/√(1-x^2) ]dx = arcsinx - √(1-x^2) + C
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