Monday

September 1, 2014

September 1, 2014

Posted by **Mariam** on Monday, June 16, 2014 at 3:42pm.

- Calculus -
**Mariam**, Monday, June 16, 2014 at 3:43pmFind the following integral ? How , is it by substitution !!!!!!!!

- Calculus - not sure how but rewrote -
**Damon**, Monday, June 16, 2014 at 4:06pmI do not know but suspect you mean

∫1/(x^2 − 2x + 8)^(3/2) dx

- Calculus -
**Damon**, Monday, June 16, 2014 at 4:08pm∫(x2 − 2x + 8)^-1.5 dx

- Calculus -
**Damon**, Monday, June 16, 2014 at 4:11pmhttp://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=7c9206fea2b96c1f22626b03b18327b3&title=Math+Help+Boards:+Indefinite+Integral+Calculator&theme=blue

says:

(x-1)/[7 sqrt(x^2-2x+8) ]

- Calculus -
**Steve**, Monday, June 16, 2014 at 4:45pmNote 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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