Question

What is the integral of (x^2+x+1)/(x^3+3x)? I know I need to use partial fractions but beyond that I'm a little lost.

Answer 1

yeah this is going to suck because the denominator has an irreducible quadratic factor
\[\frac{x^2+x+1}{x(x^2+3)}\]

Answer 2

you need all the gory details? or just the answer?

Answer 3

\[\frac{x^2+x+1}{x(x^2+3)}=\frac{a}{x}+\frac{bx+c}{x^2+3}\] we get
\[a(x^2+3)+(bx+c)x=x^2+x+1\] you can get \(a\) right away by replacing \(x\) by \(0\) to get
\[3a=1\] so \(a=\frac{1}{3}\)

Answer 4

now we have
\[ax^2+3a+bx^2+cx=x^2+x+1\] and \(a=\frac{1}{3}\) so it is
\[\frac{1}{3}x^2+bx^2+cx+1=x^2+x+1\]
since \(cx=x\) we know \(c=1\) and also since
\[\frac{1}{3}x^2+bx^2=x^2\] we have \(b=\frac{2}{3}\)

Answer 5

so we get
\[\frac{1}{3x}+\frac{2x+1}{3(x^2+3)}=\frac{1}{3}\left(\frac{1}{x}+\frac{2x+1}{x^2+3}\right)\] and now the integral should be easy

Answer 6

ok thanks, when I seperate the second part and have 1/x^2+3 would that turn out to be arctan (x/sqrt3)/sqrt3?