So here we have the integral:
int 1/(x^2-2x+2)^2 dx
And the form of quadratic reciprocal seems to suggest that trigonometric substitution would work here. So first complete the square to get:
x^2-2x+2 = (x-1)^2 +1
Then apply the substitution u = x-1 to remove the linear:
(du)/dx = 1
rArr du = dx
So we can safely change variables with no unwanted side-effects:
int 1/(x^2-2x+2)^2 dx
= int 1/((x-1)^2 +1)^2 dx
-= int 1/(u^2 + 1)^2 du
Now, this is the ideal form for executing a trigonometric substitution; u^2 + 1 suggests the Pythagorean Identity 1 + tan^2theta = sec^2theta, so we apply the substitution u = tantheta to simplify the denominator:
(du)/(d theta) = sec^2 theta
rArr du = sec^2 theta d theta
So the integral becomes:
int 1/(sec^2 theta)^2 * sec^2 theta d theta
= int 1/(sec^2 theta) d theta
-= int cos^2 theta d theta
Now, we use the double-angle formula for cos to make this antiderivative more manageable:
cos(2theta) = 2cos^2 theta - 1
hArr cos^2 theta = 1/2(cos(2 theta) + 1)
Then put that into the integral:
1/2 int cos(2 theta) + 1 d theta
=1/2 ( theta + 1/2 sin(2 theta)) + c (and reopening this with the double-angle formula for sin)
=1/2 theta + 1/2sinthetacostheta + c
Now, x-1 = u = tan theta
rArr theta = arctan(x-1)
1 + (x-1)^2 = sec^2 theta
rArr cos theta = 1/sqrt(x^2 - 2x +2)
sin theta = tan theta*cos theta
rArr sin theta = (x-1)/(sqrt(x^2+2x+2)
:. sintheta*costheta = (x-1)/(x^2-2x+2)
Finally, getting to the point:
int 1/(x^2-2x+2)^2 dx
= 1/2arctan(x-1) + (x-1)/(2(x^2-2x+2)) + c