（x^3+x-1）/（x^2+2）^2的不定积分怎么求？
（x^3+x-1）/（x^2+2）^2的不定积分怎么求？
x^3+x-1=x^3+2x-x-1
原式=∫(1/x^2+1)dx-0.5∫(d(x∧2+2)/(x∧2+2)∧2)-∫dx/(x∧2+2)∧2
1/(x^2+2)^2 的不定积分
∫1/(x^2+2)^2 dx
let
x =√2tana
dx = √2(seca)^2 da
∫1/(x^2+2)^2 dx
=∫(1/[4(seca)^4]) √2(seca)^2 da
= (√2/4)∫ (cosa)^2da
=(√2/8)∫ (cos2a+1) da
=(√2/8)[ sin2a/2 + a] + C
=(√2/8)[ √2x/(4+x^2) + arctan(x/√2)] + C
(x∧3＋ 1)/(x∧2＋ 1)∧2的不定积分怎么求
∫(x³+1)/(x²+1)² dx
= ∫(1-x)/(x²+1)² dx + ∫x/(x²+1) dx
= J + (1/2)ln(x²+1)
令x=tany,dx=sec²y dy,siny=x/√(x²+1),cosy=1/√(x²+1)
J = ∫(1-tany)/sec⁴y * sec²y dy
= ∫(1-tany)cos²y dy
= ∫cos²y dy - ∫sinycosy dy
= (1/2)∫(1+cos2y) - (1/2)∫sin2y dy
= y/2 + 1/4*sin2y + 1/4*cos2y
= (1/2)arctanx + (1/2)*x/(x²+1) + 1/4*[2/(x²+1)-1]
= (1/2)arctanx + x/[2(x²+1)] + (1-x²)/[2(x²+1)]
原积分= (1/2)arctanx + x/[2(x²+1)] + (1-x²)/[2(x²+1)] + (1/2)ln(x²+1) + C
= (1/2)[(x+1)/(x²+1) + ln(x²+1) + arctanx] + C
求x^2/（1-x^2）^3的不定积分
令x = sinz, dx = cosz dz
∫ x²/(1 - x²)³ dx
= ∫ sin²z/cos⁶z * cosz dz
= ∫ sin²z/cos⁵z dz
= - ∫ sinz/cos⁵z d(cosz)
= (- 1/4)∫ sinz d(1/cos⁴z)
= (- 1/4)(sinz/cos⁴z) + (1/4)∫ 1/cos⁴z d(sinz)
= (- 1/4)(sinz/cos⁴z) + (1/4)∫ 1/cos³z dz
= (- 1/4)(sinz/cos⁴z) + (1/4)∫ sec³z dz
= (- 1/4)(sinz/cos⁴z) + (1/4)(1/2)[secztanz + ln|secz + tanz|] + C
sinz = x ==> cosz = √(1 - x²)、secz = 1/√(1 - x²)、tanz = x/√(1 - x²)，回代就好了
求(1+3x^2)/x^2*(x^2+1)的不定积分
∫ { (1+3x^2)/[x^2*(x^2+1) } dx
=∫ {3/(x^2+1) + 1/[x^2.(x^2+1)] } dx
=3arctanx + ∫dx/[x^2.(x^2+1)]
=3arctanx + ∫ [1/x^2 -1/(x^2-1) ]dx
=3arctanx -1/x - ∫ 1/(x^2-1) dx
let
x= secy
dx= secy tany dy
∫ 1/(x^2-1) dx
=∫ (secy/tany) dy
=∫ cscy dy
=-ln|cscy -coty | + C'
=-ln|x/ √(x^2-1) - 1/√(x^2-1) | + C'
∫ { (1+3x^2)/[x^2*(x^2+1) } dx
=3arctanx -1/x - ∫ 1/(x^2-1) dx
=3arctanx -1/x +ln|x/ √(x^2-1) - 1/√(x^2-1) | + C
∫(x-1)/(x^2 2x 3)dx的不定积分怎么求
∫(x-1)/(x²+2x+3)dx
=½∫(2x-2)/(x²+2x+3)dx
=½∫(2x+2-4)/(x²+2x+3)dx
=½∫(2x+2)/(x²+2x+3)dx - ½∫4/(x²+2x+3)dx
=½∫(2x+2)/(x²+2x+3)dx - 2∫1/(x²+2x+3)dx
=½∫d(x²+2x+3)/(x²+2x+3) - 2∫1/[(x+1)²+2]dx
=½ln|x²+2x+3| - ∫1/{[(x+1)/√2]²+1}dx + C
=½ln|x²+2x+3| - (√2)∫1/{[(x+1)/√2]²+1}d[(x+1)/√2] + C
=½ln|x²+2x+3| - (√2)arctan[(x+1)/√2] + C
求 X^3+X^2+1/X^3+X^2 的不定积分
每一项都是x^a的形式，它们的原函式具有相同的形式：1／（a+1）x^（a+1）
1/（x^2-2x-3）的不定积分怎么求?
∫[1/(x²-2x-3)]dx
=∫[1/(x+1)(x-3)]dx
=¼∫[(x+1)-(x-3)]/[(x+1)(x-3)] dx
=¼∫[1/(x-3) -1/(x+1)]dx
=¼∫[1/(x-3)]d(x-3) -¼∫[1/(x+1)]d(x+1)
=¼ln|x-3|-¼|ln(x+1)|+C
=¼ln|(x-3)/(x+1)| +C
(x^2)/[(x^2+1)^(1/2)]的不定积分怎么求?
原式化简下=（x^2+1)^(1/2)-1/（x^2+1)^(1/2)
原式不定积分=∫（x^2+1)^(1/2)dx-∫1/（x^2+1)^(1/2)dx
∫（x^2+1)^(1/2)dx= x*（x^2+1)^(1/2)-∫xd（x^2+1)^(1/2)=x*（x^2+1)^(1/2)-∫x^2/（x^2+1)^(1/2)dx=x*（x^2+1)^(1/2)-原式+∫1/（x^2+1)^(1/2)dx
原式=1/2（x*（x^2+1)^(1/2)+∫1/（x^2+1)^(1/2)dx）
∫1/（x^2+1)^(1/2)dx=ln|x+(x^2+1)^(1/2)|+c
所以原不定积分就=1/2【x*（x^2+1)^(1/2)-ln|x+(x^2+1)^(1/2)|】+C
