Question

求解微积分12.13题

求解微积分12.13题12.13题

Answer 1

(12)
∫ (x+1)^2/(x^2+1)^2 dx
=∫ dx/(x^2+1) + ∫ 2x/(x^2+1)^2 dx
=arctanx - 1/(x^2+1) + C
(13)
consider
x^2+x+1 = (x+1/2)^2 + 3/4
let
x+1/2 = (√3/2)tanu
dx=(√3/2)(secu)^2 du
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∫ (-x^2-2)/(x^2+x+1)^2 dx
=-∫dx/(x^2+x+1) + (1/2)∫ (2x+1)/(x^2+x+1)^2 dx - (3/2)∫dx/(x^2+x+1)^2
=-∫dx/(x^2+x+1) - (3/2)∫dx/(x^2+x+1)^2 - (1/2) [1/(x^2+x+1)]
=-∫(√3/2)(secu)^2 du/[(3/4)(secu)^2] - (3/2)∫(√3/2)(secu)^2 du/[ (9/16)(secu)^4]
- (1/2) [1/(x^2+x+1)]
=-(2√3/3)∫ du - (4√3/3)∫ (cosu)^2 du - (1/2) [1/(x^2+x+1)]
=-(2√3/3)u -(2√3/3)∫ (1+cos2u) du - (1/2) [1/(x^2+x+1)]
=-(2√3/3)u -(2√3/3) [u+(1/2)sin2u] - (1/2) [1/(x^2+x+1)] +C
=-(4√3/3)u -(2√3/3) [(1/2)sin2u] - (1/2) [1/(x^2+x+1)] +C
=-(4√3/3)arctan[(2x+1)/√3] -(1/2)[(2x+1)/(x^2+x+1)] - (1/2) [1/(x^2+x+1)] +C
=-(4√3/3)arctan[(2x+1)/√3] -(x+1)/(x^2+x+1) +C