1. Integration

I need a bit of help integrating

X^2/(X^2-4X+5)

2. Originally Posted by qwerty10
I need a bit of help integrating

X^2/(X^2-4X+5)

Hint:

$\frac{x^2}{x^2-4x+5}=\frac{4x-8}{x^2-4x+5}+\frac{3}{x^2-4x+5}+1$

3. 1. Do a long division.
2. Apply:

$\int \dfrac{f'(x)}{f(x)} dx = ln |f(x)| + C$

4. Thank you
Is the correct way to continue by

let u=x^2-4x+5
du=(2x-4)dx

so (x-2)dx=du/4

After this I am not quite sure what to do

5. $\int \frac{x^2}{x^2-4x+5} dx = \int 1+ \frac{4x-5}{x^2-4x+5} dx = \int 1 dx + \int \frac{4x-5}{x^2-4x+5} dx$

$= \int 1 dx + \int \frac{4x-8}{x^2-4x+5} dx + \int \frac{3}{x^2-4x+5} dx$

$=x + 2ln |x^2-4x+5| + \int \frac{3}{x^2-4x+5} dx$

Can you continue now?

6. Originally Posted by qwerty10
Thank you
Is the correct way to continue by

let u=x^2-4x+5
du=(2x-4)dx

so (x-2)dx=du/4 Not Needed rather dx = du/(2x-4)

After this I am not quite sure what to do
Look at the reply by Also sprach Zarathustra

$\int \frac{x^2}{x^2-4x+5}dx= \int \frac{4x-8}{x^2-4x+5}dx+\int \frac{3}{x^2-4x+5}dx + \int 1 dx$

use the substitution rule for the first term only:

$\int \frac{4x-8}{x^2-4x+5}dx = \int \frac{2(x-4)}{x^2-4x+5}$

sub. rule gives you:

$\int \frac{2}{u} du$ . complete it..

For the second term: $\int \frac{3}{x^2-4x+5}dx = \int \frac{3}{1+(x-2)^{2}} dx$

you should be able to finish it off now..

7. Thank you

I have worked the answer out to be 2ln(x^2-4x+5) +3tan^-1(x-2) + x+C ??
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