Question

How do I find the indefinite integral of: (3x-1)/(x^2-x+1)? Thanks.

Answer 1

\[\int\limits_{}\frac{ 3x-1 }{ x^2-x+1 }dx{}\]

Answer 2

You're gonna wanna "decompose" that fraction in the integrand.
\[ \frac{ ax+b }{ (x-r)(x-t) } = \frac{ c }{ (x-r) } + \frac{ d }{ (x-t) }\]
where a,b,r,t are known numbers from the original fraction; and c,d are unknowns.

Working through the algebra you get (the dx and dr are not differentials!)
\[cx+ct+dx+dr = ax+b\]
This only holds up when all the x terms are equal and all the constant terms are equal, so we can break this into 2 equations.
\[c + d = a\]
\[ct + dr = b\]
Some algebra later and you get
\[d = \frac{ b-at }{ r-t }\]
\[c=\frac{ ar-b }{ r-t }\]
That should help you simplify your integrand. From there its just a straightforward u sub to evaluate the integral. Let me know if you need anymore help.

Answer 3

Partial fraction decomp may be a bit tricky here, as there are no real roots to the denominator's polynomial.

I would suggest the following:
\[\int\frac{3x-1}{x^2-x+1}~dx=\int\frac{x}{x^2-x+1}~dx+\int\frac{2x-1}{x^2-x+1}~dx\]
A simple \(u\)-sub works for the last integral: \(u=x^2-x+1\), so \(du=(2x-1)~dx\).

For the first integral, you would complete the square in the denominator:
\[\begin{align*}x^2-x+1&=x^2-x+\frac{1}{4}+\frac{3}{4}\\\\
&=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}
\end{align*}\]
Next step involves a preliminary algebraic sub: \(x-\dfrac{1}{2}=\dfrac{\sqrt3}{2}\tan t\), so that \(dx=\dfrac{\sqrt3}{2}\sec^2t~dt\).
\[\begin{align*}\int\frac{x}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}~dx&=\int\frac{\frac{\sqrt3}{2}\tan t+\frac{1}{2}}{\left(\frac{\sqrt3}{2}\tan t\right)^2+\frac{3}{4}}\times\dfrac{\sqrt3}{2}\sec^2t~dt\\
&=\int\frac{\frac{\sqrt3}{2}\tan t+\frac{1}{2}}{\frac{3}{4}\tan^2t+\frac{3}{4}}\times\dfrac{\sqrt3}{2}\sec^2t~dt\\
&=\frac{\frac{\sqrt3}{2}}{\frac{3}{4}}\int\frac{\frac{\sqrt3}{2}\tan t+\frac{1}{2}}{\tan^2t+1}\times\sec^2t~dt\\
&=\frac{2\sqrt3}{3}\int\frac{\frac{\sqrt3}{2}\tan t+\frac{1}{2}}{\sec^2t}\times\sec^2t~dt\\
&=\frac{2\sqrt3}{3}\int\left(\frac{\sqrt3}{2}\tan t+\frac{1}{2}\right)~dt\\
&=\int\tan t~dt+\frac{1}{2}\int dt\\\\\\
&=\cdots
\end{align*}\]

Answer 4

Thanks! Any reason I cant see the equations? all i see is code ("frac\"..)

Answer 5

Likely a browser problem? Try switching your browser, or checking back later. I'm sure you can understand why I'd be reluctant to recreate that as a drawing haha