Question

∫(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|]

Answer 1

∫[√(x²-a²)]dx=？
解：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]。