thomas5267 Group Title Rewrite $$\int^3_2\frac{3x}{x^2-4x+5}\,dx$$ into the form of $$\int^b_a\frac{mt+c}{t^2+k^2}$$ or $$\int^b_a\frac{mt+c}{t^2-k^2}$$. Completing the square, as far as I can see, wouldn't work this time. What should I do with this? one year ago one year ago

1. Callisto Group Title

It works :)

2. Callisto Group Title

$\int^3_2\frac{3x}{x^2-4x+5}dx=\int^3_2\frac{3x}{x^2-4x+(4+1)}dx=\int^3_2\frac{3x}{(x-2)^2+1}dx$In this case, t=x-2, k=1

3. thomas5267 Group Title

*facepalm* I am such an idiot.

4. Callisto Group Title

No, you're not..