Question

一道不定积分题

这个不定积分怎么求？

Answer 1

解：设x=√atanθ，则dx=√asec²θdθ,sinθ=x/√(x²+a),cosθ=√a/√(x²+a)
∴原式=∫√asecθ√asec²θdθ
=a∫cosθ/(cosθ)^4dθ
=a∫d(sinθ)/(1-sin²θ)²
=a/4∫[(2+sinθ)/(1+sinθ)²+(2-sinθ)/(1-sinθ)²]d(sinθ)
=a/4∫[1/(1+sinθ)²+1/(1+sinθ)+1/(1-sinθ)²+1/(1-sinθ)]d(sinθ)
=a/4[ln(1+sinθ)-ln(1-sinθ)-1/(1+sinθ)+1/(1-sinθ)]+C1 (C1是积分常数)
=a/4{ln[(1+sinθ)/(1-sinθ)]+2sinθ/(1-sin²θ)}+C1
=a/4{2ln[x+√(x²+a)]-lna+2x√(x²+a)}+C1
=a/2{ln[x+√(x²+a)]+x√(x²+a)}-alna/4+C1
=a/2{ln[x+√(x²+a)]+x√(x²+a)}+C (C=C1-alna/4,C也是积分常数)。

Answer 2

换元，令x=根号下a乘以sht