∫dx/(1+x)(1+x^2)=?
∫dx/(1+x)(1+x^2)=?
可用待定系数法。
令1/[(1 + x)(1 + x^2)] = A/(1 + x) + (Bx + C)/(1 + x^2)
==> 1 = A(1 + x^2) + (Bx + C)(1 + x)
==> 1 = (A + B)x^2 + (B + C)x + (A + C)
∴A + B = 0 ==> B = - A
∴B + C = 0 ==> C = - B
∴A + C = 1 ==> C = 1 - A
有1 - A = - (- A) ==> A = 1/2、B = - 1/2、C = 1/2
於是∫ 1/[(1 + x)(1 + x^2)] dx
= (1/2)∫ 1/(1 + x) dx - (1/2)∫ x/(1 + x^2) dx + (1/2)∫ 1/(1 + x^2) dx
= (1/2)ln|1 + x| - (1/4)ln(1 + x^2) + (1/2)arctan(x) + C
= (1/4)ln[(1 + x)^2/(1 + x^2)] + (1/2)arctan(x) + C
∫dx/x^2√(1+x^2)
F(x)=∫dx/[(x^2)*((1+x^2)^(1/2))]
设x=tant,则dx=sec²tdt,
∴F(x)=∫dx/[(x^2)*((1+x^2)^(1/2))]
=∫sec²tdt/[tan²t*(1+tan²t)^(1/2)]
=∫sec²tdt/(tan²t*sect)
=∫sectdt/tan²t
=∫(cos²t/sin²t)*(1/cost)*dt
=∫(cost/sin²t)dt
=∫dsint/sin²t
=(-1/sint) +C
∫dx/x^2(1+x^2)
原式=∫[1/x²-1/(1+x²)]dx
=-1/x-arctanx+C
∫dx/(x^4)(1+x^2)^(1/2)
令x=tanu,则dx=sec²udu,(1+x^2)^(1/2)=secu
原式=∫ sec²u/[(tanu)^4secu] du
=∫ sec²u/[(tanu)^4secu] du
=∫ secu/(tanu)^4 du
=∫ cos³u/(sinu)^4 du
=∫ cos²u/(sinu)^4 d(sinu)
=∫ (1-sin²u)/(sinu)^4 d(sinu)
=∫ 1/(sinu)^4 d(sinu) - ∫ 1/sin²u d(sinu)
=-(1/3)(sinu)^(-3) + 1/sinu + C sinu=x/√(1+x²)
=-(1/3)(1+x²)^(3/2)/x³ + √(1+x²)/x + C
希望可以帮到你,不明白可以追问,如果解决了问题,请点下面的"选为满意回答"按钮,谢谢。
∫(+∞,1)dx/1+x^2=?
1/1+x^2的不定积分=arctanx
然後带入(这个是广义的不定积分哦)
=arctan(+∞)-arctan1
=π/2-π/4
=π/4
希望你明白了O(∩_∩)O哈!
(如果没有记错,你昨天好像问过我这个题目,因为昨天做了一个这个题目)
∫dx/(1+x^2)(1+arctan^2x)=
∫dx/[ (1+x^2)(1+(arctanx)^2)]
let
y=arctanx
dy =[ 1/(1+x^2)] dx
∫dx/[ (1+x^2)(1+(arctanx)^2)]
=∫dy/(1+y^2)
=arctany + C
=arctan((arctanx))+ C
∫dx/(1+x^2)^3
令x=tant 则,则dx=(sect)^2dt
∫dx/(1+x^2)^3=
∫(sect)^2*1/(sect)^6 dt
=∫(cost)^4dt
=∫(1+cos2t)^2/4dt
=1/4∫(1+2cos2t+(cos2t)^2)dt
=1/4∫(1+2cos2t+1/2(1+cos4t))dt
=3/8t+1/4sin2t+1/32sin4t
反带回去!
=3/8arctanx+1/2x√(1-x^2)+1/8(x-2x^2)√(1-x^2)
∫dx/(1+x^2)^3/2 = ?
令x=tant 则(1+x^2)^3/2=(sect)^3 dx=(sect)^2dt
∫dx/(1+x^2)^3/2 =∫costdt=sint+c=x/(1+x^2)^1/2+c
∫dx/[(1+x^2)^(3/2)]
令x=tant, -π/2<t<π/2
原式化为:
∫sec²tdt/sec³t
=∫costdt
=sint+C
=x/√(1+x²)+C
∫(x^4+2x^2+x+1)dx/[x(1+x^2)]
∫ (x^4+2x²+x+1)/[x(1+x²)] dx
=∫ (x^4+x²)/[x(1+x²)] dx + ∫ (x²+1)/[x(1+x²)] dx + ∫ x/[x(1+x²)] dx
=∫ x dx + ∫ 1/x dx + ∫ 1/(1+x²) dx
=(1/2)x² + ln|x| + arctanx + C
【数学之美】团队为您解答,若有不懂请追问,如果解决问题请点下面的“选为满意答案”。
2024-04-11 广告