Factorise 1+x2

Anthonyk2013 said:
Only done basic numbers no idea how to do this, point me right direction please
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Has your book or your instructor mentioned "prime" polynomials? They're the polys that can't be factored (at least not using just the "real" numbers). In your case, the poly is similar to "13" in the natural numbers; 13 can't be factored, either. ;)
 
stapel said:
Has your book or your instructor mentioned "prime" polynomials? They're the polys that can't be factored (at least not using just the "real" numbers). In your case, the poly is similar to "13" in the natural numbers; 13 can't be factored, either. ;)
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I got 1+x2 from a integration question im doing. I have to factorise whats under the line
 
Anthonyk2013 said:
I got 1+x2 from a integration question im doing. I have to factorise whats under the line
Click to expand...

Perhaps you don't. I suspect you are trying to use "partial fractions" but, again, you cannot factor "1+ x2 in terms of real numbers. You need to know that \(\displaystyle \int \frac{dx}{1+ x^2}= arctan(x)+ C\).


More generally, if you have a fraction of the form \(\displaystyle \frac{1}{(x- a)(x- b)(x^2+ cx+ d)}\), where \(\displaystyle x^2+ cx+ d= 0\) does not have real roots, so that it cannot be factored into linear factors, you can "complete the square", so that \(\displaystyle x^2+ cx+ d= (x- x_0)^2+ g\) and then write
\(\displaystyle \frac{A}{x- a}+ \frac{B}{x- b}+ \frac{Cx+ D}{(x- x_0)^2+ g}\)
The first two fractions can be integrated as logarithms. The third fraction can be written as \(\displaystyle \frac{C(x- x_0)+ Cx_0+ D}{(x- x_0)^2+ g}= \frac{C(x- x_0)}{(x- x_0)^2+ g}+ \frac{Cx_0+ D}{(x- x_0)+ g}\). Letting \(\displaystyle u= x- x_0\), that is \(\displaystyle \frac{Cu}{u^2+ g}+ \frac{Cx_0+ D}{u^2+ g}\) the first can be integrated by letting \(\displaystyle u^2+ g\), the second as an arctangent.
 
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