求不定积分?
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∫(2x+5)/(x^2+2x+5) dx
=∫(2x+2)/(x^2+2x+5) dx + 3∫dx/(x^2+2x+5)
=ln|x^2+2x+5| +3∫dx/(x^2+2x+5)
=ln|x^2+2x+5| +(3/2)arctan[(x+1)/2] + C
//
x^2+2x+5 =(x+1)^2 +4
let
x+1 = 2tanu
dx =2(secu)^2 du
∫dx/(x^2+2x+5)
=∫2(secu)^2 du/[ 4(secu)^2 ]
=(1/2)u + C
=(1/2)arctan[(x+1)/2] + C
=∫(2x+2)/(x^2+2x+5) dx + 3∫dx/(x^2+2x+5)
=ln|x^2+2x+5| +3∫dx/(x^2+2x+5)
=ln|x^2+2x+5| +(3/2)arctan[(x+1)/2] + C
//
x^2+2x+5 =(x+1)^2 +4
let
x+1 = 2tanu
dx =2(secu)^2 du
∫dx/(x^2+2x+5)
=∫2(secu)^2 du/[ 4(secu)^2 ]
=(1/2)u + C
=(1/2)arctan[(x+1)/2] + C
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OK多谢
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