Use the sub x-1 = sqrt(5)tan(u) to evaluate 5∫3 1/(x^2-2x+6)^3/2

Question

I don't know how to use the substitution

Answer 1

well, if tan u = (x-1)/sqrt5
then sin u= (x-1)/(x^2-2x+6)^1/2 (draw the triangle u out)
and sec u= (x^2-2x+6)^1/2 / sqrt5

and sec^2 u=(x^2-2x+6)/5

At this point you should be able now to substitute in the integral for x the u function.

Answer 2

x^2-2x+6 = (x-1)^2 + 5

if x-1 = √5 tanu then

(x-1)^2+5 = 5tan^2u+5 = 5sec^2u
and
dx = √5 sec^2u du

plug those into the integral and watch things simplify

Answer 3

Would it be possible to give me a more step by step? Im lost

Answer 4

x-1 = √5 tan(u)
x^2-2x+6 = 5sec^2(u)
dx = √5 sec^2(u) du

Now plug it into the integral

5∫3 1/(x^2-2x+6)^3/2 dx
5∫3 1/(5sec^2(u))^3/2 √5 sec^2(u) du
5∫3 (√5 sec^2(u))/(5√5 sec^3(u)) du
5∫3 1/(5 sec(u)) du
∫3 cos(u) du
3 sin(u) + C

now, if
x-1 = √5 tan(u),
tan(u) = (x-1)/√5
sin(u) = (x-1)/√(x^2-2x+6)

so the final answer is

3(x-1)/(x^2-2x+6) + C

Not sure just what the 5∫3 means, but if it's some kind of typo, I'm sure you can adjust things.

Answer 5

Oh, thank you. The 5∫3 just mean the upper and lower limts, I didn't know how else to write it