∫1/(x2 − 2x + 8)3/2 dx
5 answers
Find the following integral ? How , is it by substitution !!!!!!!!
I do not know but suspect you mean
∫1/(x^2 − 2x + 8)^(3/2) dx
∫1/(x^2 − 2x + 8)^(3/2) dx
∫(x2 − 2x + 8)^-1.5 dx
http://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) ]
says:
(x-1)/[7 sqrt(x^2-2x+8) ]
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.
∫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.