Integral 0 to 1 1/(1+sqrtx)^4 dx
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Integral 0 to 1 1/(1+sqrtx)^4 dx

[From: ] [author: ] [Date: 12-04-25] [Hit: ]
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I try substituting u=sqrtx and ended up with the integral 2 * 0 to 1 u/(1+u)4 which would take too long multiplying it all out. What's the correct substitution?

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∫dx/(1 + √x)^4 from 0 to 1

u = 1 + √x

(u - 1)² = x

2(u - 1) du = dx

2*∫(u - 1)/(u^4) du from 1 to 2

2*∫(u^(-3) - u^(-4)) du from 1 to 2

= 2*[-1/2*u^(-2) + 1/3*u^(-3) eval. from 1 to 2]

= 2[-1/8 + 1/24 + 1/2 - 1/3]

= 1/6

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No need to expand. Write u/(u+1)^4 = [1/(u+1)^3 - 1/(u+1)^4] and then integrate.
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keywords: to,sqrtx,Integral,dx,Integral 0 to 1 1/(1+sqrtx)^4 dx
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