how do you integrate this expression?

Question

given then y=1/(2x^2 - 2x +1)^(3/2)

how do you integrate y?

thanks for the help, urgently needed.

Answer 1

First write the integrand more conveniently. The denominator can be written

(2x^2 - 2x +1)^(3/2)
= 2sqrt2 [x^2 - x + 1/2]^(3/2)
= 2sqrt2[(x-1/2)^2 + 1/4]^(3/2).

Now translate u = x - 1/2, so du = dx, and the integrand is

1 / 2sqrt2(u^2 + 1/4)^(3/2).

Now use a trigonometric substitution: Set 2u = cot theta, which makes sin theta = 1 / 2sqrt(u^2 + 1/4), and

2sqrt2 (sin theta)^3 = 1 / 2sqrt2(u^2 + 1/4)^(3/2).

Now, du = -1/2 (csc theta)^2 dtheta, and we can substitute all this in. The integrand is:

(1 / 2sqrt2) -(csc theta)^2 dtheta 2sqrt2 (sin theta)^3,

which equals

(-1/2) sin theta dtheta.

Integrate this, then substitute theta = arccot(2u), then finally substitute u = x + 1/2.

There may also be another way that's faster, involving the binomial series, but this will definitely work.

Answer 2

Complete the square in the denominator.
I always find this easier if I factor out the coefficient of x^2 first, so

2x^2 - 2x +1 = 2[ x^2 - x +1/2]
= 2[ (x-1/2)^2 +1/4]

this is all raised to the 3/2 in the denominator, so bring the constant 1/2^(3/2) outside the integral. then let u= x-1/2, for which du = dx, so you have

1/2^(3/2) * Int {1/[ u^2 +1/4]^(3/2)} du

Now either use an integral table or, if you are going to do this by hand all the way, use the substitution
u = (1/2) tan t.