Question

∫(x^2+a^2)^1/2dx

我看到有人说用sht代换x会很简单，但是我还是算不出来，哪位数学帝帮帮忙。。。。
如果是三角代换就不用发了，我会了

Answer 1

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

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Answer 2

高数课本里有