# Calculus

posted by Mariam

∫1/(x2 − 2x + 8)3/2 dx

1. Mariam

Find the following integral ? How , is it by substitution !!!!!!!!

2. Damon

I do not know but suspect you mean

∫1/(x^2 − 2x + 8)^(3/2) dx

3. Damon

∫(x2 − 2x + 8)^-1.5 dx

4. Damon
5. Steve

Note that you have

∫1/(x^2 − 2x + 8)^(3/2) dx
= ∫1/((x-1)^2 + 7)^(3/2) dx
If x-1 = √7 tanu
(x-1)^2 + 7 = 7 sec^2 u
dx = √7 sec^2 u du
and you have
∫1/(√7 secu)^3 * √7 sec^2 u du
= ∫ 1/7 cosu du
= 1/7 sinu
= 1/(7 csc u)
= 1/(7 √(1+cot^2 u))
= 1/(7 √(1 + (7/(x-1)^2))
= 1/(7/(x-1) √((x-1)^2 + 7)
= 1/ 7√(x^2-2x+8)

Simpler than I'd have expected.

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