Question

求一道高数题答案：

反常积分计算∫（上限正无穷，下限0）dx/(√ (x*(x+1)^5))的值为（）
A.无穷 B.0 C.2/3 D.1

Answer 1

答案是4/3，是不是答案C错了呢？

先求不定积分：
∫dx√[x(x+1)^5]
令u=x+1
∫du/√[(u-1)u^5]
令√u=secβ
sinβ=√(u-1)/√u
u=sec²β
du=2sec²βtanβ
u-1=tan²β
原式=(2sec²βtanβ)/√[tan²β(secβ)^10] dβ
=2∫sec²β/(secβ)^5 dβ
=2∫cos³β dβ
=2∫(1-sin²β) d(sinβ)
=2sinβ-(2/3)sin³β+C
=2√(u-1)/√u-[2(u-1)^(3/2)]/[3u^(3/2)]+C
=2√x/√(x+1)-2x√x/[3(x+1)^(3/2)]+C
=[2(2x+3)√x]/[3(x+1)^(3/2)]+C

∫(0到+∞) dx/√[x(x+1)^5]
=lim(b→+∞) ∫(0到b) dx/√[x(x+1)^5]
=(2/3)lim(b→+∞) [(2b+3)√b]/[(b+1)√(b+1)]
=(2/3)lim(b→+∞) (2+3/b)/(1+1/b)*[√b/√(b+1)]
=(2/3)lim(b→+∞) (2+3/b)/(1+1/b)*1/[√(b+1)/√b]
=(2/3)lim(b→+∞) (2+3/b)/(1+1/b)*1/√(1+1/b)
=(2/3)*(2+0)/(1+0)*1/√(1+0)
=4/3

Answer 2

∫dx/(√ (x*(x+1)^5))
x=(tanu)^2 dx=[1/(cosu)^2]du
x*(x+1)^5=[(sinu)^2/(cosu)^2]*[1/cosu^10]
√(x*(x+1)^5)=cosu^6/sinu
原式=∫cosu^4du/sinu=∫(1-(sinu)^2)^2du/sinu=∫du/sinu-∫2sinudu+∫(sinu)^3du
=∫-dcosu/(1-cosu^2)-(sinu)^2+(sinu)^4/4
=(-1/2)ln[(1+cosu)/(1-cosu)] - (sinu)^2+(sinu)^4/4+C
cosu=1/√(1+x) sinu=√[x/(1+x)]
(1+cosu)/(1-cosu)=[√(x+1)+1]/[√(x+1)-1]=[√(x+1)+1]^2/x
∫dx/(√(x*(x+1)^5))=√x/[√(x+1)+1] -x/(1+x)+ x^2/[4(x+1)^2]
lim(x→+∞)√x/(√(x+1)+1)=lim(x→+∞)1/[√1+(1/x)+1/x]=1
lim(x→+∞) x/(1+x)=lim(x→+∞) 1/[(1/x)+1]=1
lim(x→+∞)x^2/[4(x+1)^2]=lim(x→+∞) 1/[4(1+1/x)^2]=1/4
∫[0,+∞]dx/(√ (x*(x+1)^5))=1/4

Answer 3

如果是u=1/√ (x*(x+1)^5))
∫（上限正无穷，下限0）u dx＝4/3

Answer 4

题目意思不是很明白，多用些括号，或则拍照片

Answer 5

刷财富！！！！