Well you know that:
so let's let u = x-2
so du= dx and u+1=x-1
Then I would do a second substitution with trigonometry:
These questions share a similar theme.
1) Integral of (x-1)dx/(x^2 - 4x + 5)
I have tried dividing this problem into two fractions, without result. I thought maybe difference of squares somehow on the bottom, x^2 - 4x + 4 + 1, but what then?
2) xdx/x^4 + x^2 + 1
I set u = x^2
du = 2x
= 1/2 Integral of du/u^2+u+1
And I STILL can't solve it! It is in a really simple form now, 1/u^2 + u + 1, what am I doing wrong?
Seems like it would be obvious, but it's evading me.
Sigh. Why is this not obvious to me? V. frustrating
Well the truth is that I went through calculus I, II and III with only power rule, product rule, derivative of sine and cosine memorized.
Something useful to remember with trig substitutions though:
divide by cos^2 on both sides:
divide by sin^2 on both sides:
And you don't have to memorize is it tan^2 -1 = csc^2 ? or whatever just straight and simple.