∫√(x²＋a²)dx怎么求
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∫√(x^2＋a^2)dx
=x√(x^2＋a^2) - ∫[ x^2/ √(x^2＋a^2) ]dx
=x√(x^2＋a^2) - ∫√(x^2＋a^2) dx + a^2.∫dx/ √(x^2＋a^2)
2∫√(x^2＋a^2)dx = x√(x^2＋a^2) + a^2.∫dx/ √(x^2＋a^2)
∫√(x^2＋a^2)dx =(1/2) [ x√(x^2＋a^2) + a^2.∫dx/ √(x^2＋a^2) ]
=(1/2) [ x√(x^2＋a^2) + a^2.ln| √(x^2＋a^2) + x | ] + C
consider
let
x=atanu
dx = a(secu)^2 du
∫dx/ √(x^2＋a^2)
=∫ secu du
=ln|secu + tanu | + C'
=ln| (1/a)√(x^2＋a^2) + x/a | + C'
=ln| √(x^2＋a^2) + x | + C''
=x√(x^2＋a^2) - ∫[ x^2/ √(x^2＋a^2) ]dx
=x√(x^2＋a^2) - ∫√(x^2＋a^2) dx + a^2.∫dx/ √(x^2＋a^2)
2∫√(x^2＋a^2)dx = x√(x^2＋a^2) + a^2.∫dx/ √(x^2＋a^2)
∫√(x^2＋a^2)dx =(1/2) [ x√(x^2＋a^2) + a^2.∫dx/ √(x^2＋a^2) ]
=(1/2) [ x√(x^2＋a^2) + a^2.ln| √(x^2＋a^2) + x | ] + C
consider
let
x=atanu
dx = a(secu)^2 du
∫dx/ √(x^2＋a^2)
=∫ secu du
=ln|secu + tanu | + C'
=ln| (1/a)√(x^2＋a^2) + x/a | + C'
=ln| √(x^2＋a^2) + x | + C''
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