高数,需详解?
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∫ (2x-7)/(4x^2+12x+25) dx
=(1/4) ∫ (8x+12)/(4x^2+12x+25) dx - 10∫ dx/(4x^2+12x+25)
=(1/4)ln|4x^2+12x+25| -10∫ dx/(4x^2+12x+25)
=(1/4)ln|4x^2+12x+25| -(5/4)arctan(2x+3) +C
consider
4x^2+12x+25
=4(x+ 3/2)^2 + 16
let
x+ 3/2 = 2tanu
dx = 2(secu)^2 du
∫ dx/(4x^2+12x+25)
=∫ 2(secu)^2 du/ [16(secu)^2 ]
=(1/8)u +C'
=(1/8)arctan(2x+3) +C'
=(1/4) ∫ (8x+12)/(4x^2+12x+25) dx - 10∫ dx/(4x^2+12x+25)
=(1/4)ln|4x^2+12x+25| -10∫ dx/(4x^2+12x+25)
=(1/4)ln|4x^2+12x+25| -(5/4)arctan(2x+3) +C
consider
4x^2+12x+25
=4(x+ 3/2)^2 + 16
let
x+ 3/2 = 2tanu
dx = 2(secu)^2 du
∫ dx/(4x^2+12x+25)
=∫ 2(secu)^2 du/ [16(secu)^2 ]
=(1/8)u +C'
=(1/8)arctan(2x+3) +C'
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