∫(x^2-a^2)^1/2dx=?要详细过程,答案是1/2[x(x^2-a^2)^1/2+a^2㏑|x+(x^2-a^2)^1/2|]

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∫(x^2-a^2)^1/2dx=?要详细过程,答案是1/2[x(x^2-a^2)^1/2+a^2㏑|x+(x^2-a^2)^1/2|]
∫(x^2-a^2)^1/2dx=?要详细过程,答案是1/2[x(x^2-a^2)^1/2+a^2㏑|x+(x^2-a^2)^1/2|]

∫(x^2-a^2)^1/2dx=?要详细过程,答案是1/2[x(x^2-a^2)^1/2+a^2㏑|x+(x^2-a^2)^1/2|]
∫[√(x²-a²)]dx=?
设x=asecu,则dx=asecutanudu,x²-a²=a²(sec²u-1)=a²tan²u,√(x²-a²)=atanu,
secu=x/a,tanu=[√(x²-a²)]/a.
代入原式得：∫[√(x²-a²)]dx=a²∫tan²usecudu=a²∫secu(sec²u-1)du=a²[∫sec³udu-∫secudu]
=a²[(1/2)secutanu+(1/2)ln(secu+tanu)-ln(secu+tanu)]+lnc₁
=a²[(1/2)secutanu-(1/2)ln(secu+tanu)]+lnc₁
=(1/2)[(x√(x²-a²)]-(a²/2)ln[(x/a)+(1/a)√(x²-a²)]+lnc₁
=(1/2){x√(x²-a²)-a²[ln(x+√(x²-a²)-lna]}+lnc₁
=(1/2){x√(x²-a²)-a²ln[x+√(x²-a²)]}+a²ln(ac₁)
=(1/2){x√(x²-a²)-a²ln[x+√(x²-a²)]}+C
[原答案错个符号].

∫x[x/[(2a-x)]^(1/2)dx=? ∫（1／a^2-x^2)dx 求∫ (dx / a^2- x^2) （a>0常数）附加个：∫ (dx / (a-x)(a+x))= 1/2a∫ ((a-x)+(a+x) / (a-x)(a+x))dx 这是怎么换算的? ∫ x/(1+X^2)dx= ∫(x+1/x)^2dx=? ∫(x^2+a^2)^(-1/2)dx=? 下列无穷积分收敛的是 A ∫sinx dx B ∫e^-2x dx C ∫1/x dx D∫1/√x dx ∫[dx/(e^x(1+e^2x)]dx ∫1/x√(a^2-x^2)dx ∫(x-1)^2dx, ∫x^1/2dx ∫(1+x)/(X^2)dx=∫ [(1+x)/(X^2)]dx得什么? ∫dx/(x^2-a^2)只会做到这一步∫1/(x-a)(x+a)dx ∫1/1-x^2dx= ∫sqr(a^2+x^2)dx ∫(1+x^2)dx = ∫(2cosx +1／x)dx= ∫（lnx)/(1+x^2)dx=?