根号下4加x平方的原函数是多少
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I=∫√(x^2+4)dx,设x=2tant,则:
I=∫√(4tan^2t+4)d2tant
=∫2sect*2sec^2tdt
=4∫dt/cos^3t
=4∫costdt/(1-sin^4 t)
=4∫dsint/(1-sin^4 t)
=2∫dsint/(1-sin^2 t)+2∫dsint/(1+sin^2 t)
=ln[(1+sint)/(1-sint)]+2arctan(sint)+c
=2ln[x+√(x^2+4)]+2arctan[x/√(x^2+4)]-ln4+c
=2ln[x+√(x^2+4)]+2arctan[x/√(x^2+4)]+C.
I=∫√(4tan^2t+4)d2tant
=∫2sect*2sec^2tdt
=4∫dt/cos^3t
=4∫costdt/(1-sin^4 t)
=4∫dsint/(1-sin^4 t)
=2∫dsint/(1-sin^2 t)+2∫dsint/(1+sin^2 t)
=ln[(1+sint)/(1-sint)]+2arctan(sint)+c
=2ln[x+√(x^2+4)]+2arctan[x/√(x^2+4)]-ln4+c
=2ln[x+√(x^2+4)]+2arctan[x/√(x^2+4)]+C.
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