Integration of algebraic fractions

By Anonymous on Tuesday, May 1, 2001 - 09:03 pm :

This is the last part of a question I was trying to do. Whats the best way to do this integration?

(x+2)/(x^2+2x+2)+(2-x)/(x^2-2x+2) dx
Thanks.

By Kerwin Hui (Kwkh2) on Tuesday, May 1, 2001 - 10:49 pm :

Rewrite the integral as

\int \frac{1}{2} \frac{2 x + 2}{x^{2} + 2 x + 2} + \frac{1}{(x + 1)^{2} + 1} - \frac{1}{2} \frac{2 x - 2}{x^{2} - 2 x + 2} + \frac{1}{(x - 1)^{2} + 1} dx

and proceed to use your standard formulae for integration.

Kerwin

By Anonymous on Tuesday, May 1, 2001 - 10:58 pm :

It looks very nice Kerwin and I am sure I could reduce that into its standard integrals.

But I am unsure of how you got it to the form you show me.

I can get it from the original form to the form below -

$\int (x + 2) / [(x + 1)^{2} + 1] + (2 - x) / [(x - 1)^{2} + 1]$
What do we do now to get it into the form you have it.

By Kerwin Hui (Kwkh2) on Tuesday, May 1, 2001 - 11:01 pm :

Just break the numerator of both fractions:

$x + 2 \equiv (x + 1) + 1 \equiv \frac{1}{2} (2 x + 2) + 1$

$2 - x \equiv (1 - x) + 1 \equiv - \frac{1}{2} (2 x - 2) + 1$

and complete the square on the denominators:

$x^{2} + 2 x + 2 \equiv (x^{2} + 2 x + 1) + 1 \equiv (x + 1)^{2} + 1$

$x^{2} - 2 x + 2 \equiv (x^{2} - 2 x + 1) + 1 \equiv (x - 1)^{2} + 1$

Kerwin

By Anonymous on Tuesday, May 1, 2001 - 11:11 pm :

Thanks Kerwin.

I am not sure whether I could have deduced that from scratch. It is something we are expected to know and do? Or is there another way to do it?

By The Editor :

The above method is one of a number of strategies for dealing with integration of algebraic fractions. Here's a list of things that can be helpful for different algebraic fractions.

Recognising it as a standard integral eg of sin^-1 or tan^-1
Observing that the numerator is (a multiple of) the derivative of the denominator: integral of f'(x)/f(x) dx = ln|f(x)|+c
1/(ax+b)ⁿ is straightforward by substituting u=ax+b
Separating it into two fractions by breaking the numerator as above: you probably noticed that Kerwin chose how to do it to give himself nice easy fractions that could be integrated by the first two methods listed here
Separating it into two (or more) fractions by using partial fractions.

By Brad Rodgers (P1930) on Wednesday, May 2, 2001 - 03:39 am :

It is nearly worth noting that the integral of (mx+n)/(ax² +bx+c) is

$\frac{m \ln (a x^{2} + b x + c)}{2 a} + \frac{2 n - m \frac{b}{a}}{\sqrt{4 b c - c^{2}}} \tan^{- 1} (\frac{2 a x + b}{\sqrt{4 b c - c^{2}}})$

although I don't think anyone would commit this to memory.

Brad

By Anonymous on Thursday, May 3, 2001 - 12:07 am :

Thanks for that Brad.
Though I doubt whether my memory can recall that with ease. But it's well worth taking note of. Thanks again