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∫1/√(x^2+1)Dx=?

令x = tany,dx = sec²y dy,y∈(- π/2,π/2) ∫ 1/√(1 + x²) dx = ∫ 1/√(1 + tan²y) * sec²y dy = ∫ 1/|secy| * sec²y dy = ∫ secy dy,在y∈(- π/2,π/2)上secy > 0 = ln| secy + tany | + C = ln| tany + √(1 + tan...

我想你的题应该是∫1/(x+√(1-x²))dx吧? 令x=sinu,√(1-x²)=cosu,dx=cosudu ∫1/(x+√(1-x²))dx =∫1/(sinu+cosu)*(cosu)du =∫cosu/(sinu+cosu)du =1/2∫(cosu+sinu+cosu-sinu)/(sinu+cosu)du =1/2∫(cosu+sinu)/(sinu+cosu)du+1/2∫(co...

∫1/[x√(x^2-1)]dx=∫(1/x^2)/[√(x^2-1)/x]dx=∫(1/x^2)dx/√[1-(1/x)^2]=-∫d(1/x)/√[1-(1/x)^2]=-arcsin(1/x)+C其中C为任意常数∫1/[x√(1-x²)]dx分子分母同乘以x=∫x/[x²√(1-x²)]dx=(1/2)∫1/[x²√(1-x²)]d(x²)令√(1-x&#...

解: 令x=secu,则u=arcsecx ∫[1/√(x²-1)]dx =∫[1/√(sec²u-1)]d(secu) =∫(secu·tanu/tanu)du =∫secudu =ln|secu +tanu| +C =ln|x+√(x²-1)| +C

进行凑微分即可 得到∫√(1+x^2) xdx =1/2 *∫√(1+x^2) dx^2 =1/2 * 2/3 *(1+x^2)^(3/2) +C =1/3 *(1+x^2)^(3/2) +C,C为常数

x=tant ∫1/[x√(x²+1)]dx=∫1/[tant√(tan²t+1)]dtant =∫1/sintdt=-∫1/sin²tdcost=-∫1/(1-cos²t)dcost =-1/2∫1/(1-cost)+1/(1+cost)dcost =1/2ln[(1-cost)/(1+cost)}+C =ln|√(1/tan²t+1)-1/tant|+C =ln|√(1/x²+1)-1/...

x=tant ∫1/[x√(x²+1)]dx=∫1/[tant√(tan²t+1)]dtant =∫1/sintdt=-∫1/sin²tdcost=-∫1/(1-cos²t)dcost =-1/2∫1/(1-cost)+1/(1+cost)dcost =1/2ln[(1-cost)/(1+cost)}+C =ln|√(1/tan²t+1)-1/tant|+C =ln|√(1/x²+1)-1/...

凑微分法 ∫x/√(1-x^2)dx =-1/2∫d(1-x^2)/√(1-x^2) =-1/2∫[(1-x^2)^(-1/2)]d(1-x^2) =-1/2*2*(1-x^2)^(1/2)+C = -√(1-x^2)+C

∫√(1+x²) dx=√(1+x²) *x-∫x*d√(1+x²) =√(1+x²) *x-∫x*x/√(1+x²)dx=√(1+x²) *x-∫(x²+1-1)/√(1+x²)dx=√(1+x²) *x-∫[√(x²+1)-1/√(1+x²)]dx=√(1+x²) *x-∫√(x²+1)dx+∫1/√(1+x&sup...

∫1/[x+√(1+x²)]dx=∫√(1+x²)-xdx =2√(1+x²)³/3-x²/2+C

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