Question

高数 不定积分 求详解

高数 不定积分 求详解求详解啊

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Answer 1

∫√(x^2+2x+2) /x dx

=∫ (x^2+2x+2)/ [x √(x^2+2x+2) ] dx
=∫ ( x + 2 + 2/x ) /√(x^2+2x+2) dx
=(1/2) ∫ ( 2x + 2) /√(x^2+2x+2) dx +∫ dx /√(x^2+2x+2) +2∫ dx/[x.√(x^2+2x+2) ]
=√(x^2+2x+2) +∫ dx /√(x^2+2x+2) +2∫ dx/[x.√(x^2+2x+2) ]
=√(x^2+2x+2) +ln| √(x^2+x+2) + x+1 | +(2/√2)ln|√2.√(x^2+2x+2)/x - (x+2)/x | +C
/
x^2+2x+2 = (x+1)^2 +1

let
x+1 = tanu
dx = (secu)^2 du
∫ dx /√(x^2+2x+2)
=∫ (secu)^2 du / secu
=∫ secu du
=ln|secu + tanu | +C
=ln| √(x^2+x+2) + x+1 | +C1
/
∫ dx/[x.√(x^2+2x+2) ]
=∫ (secu)^2 du / [(tanu-1) secu ]
=∫ [secu / (tanu-1)] du
=∫ du/ (sinu-cosu)
=∫ du/ [√2sin(u-π/4) ]
=(1/√2) ln | csc(u-π/4) - cot(u-π/4) | +C2
=(1/√2) ln | √2.√(x^2+2x+2)/x - (x+2)/x | +C2
/
x+1 = tanu
sinu =( x+1)/√(x^2+2x+2)
cosu =1/√(x^2+2x+2)
csc(u-π/4)
=1/sin(u-π/4)
=√2/[ ( x+1)/√(x^2+2x+2) -1/√(x^2+2x+2) ]
=√2. √(x^2+2x+2)/x
cos(u-π/4) = (x+2)/[√2. √(x^2+2x+2) ]
cot(u-π/4)

= cos(u-π/4)/sin(u-π/4)
=(x+2)/x