1. ## Rational Function Integration

Find the indefinite integral of the function:

2. Doing this on Mathematica shows the answer has three separate terms which makes me think that partial fractions will be involved. I would try factoring the denominator and see where that leads.

The denominator factors to: $\displaystyle (x-1)(x^2-2x+10)$.

Thus you have $\displaystyle \frac{2x-29}{(x-1)(x^2-2x+10)}$.

Partial fractions: $\displaystyle \frac{2x-29}{(x-1)(x^2-2x+10)} = \frac{A}{x-1}+\frac{Bx+C}{x^2-2x+10}$

Eliminating the denominators yields $\displaystyle 2x-29 = A(x^2-2x+10) + (Bx+C)(x-1)$.

Multiplying this out gives $\displaystyle 2x-29 = (A+B)x^2 + (-2A-B+C)x + (10A-C)$.

Set $\displaystyle A+B = 0$, $\displaystyle -2A-B+C=2$, and $\displaystyle 10A-C=-29$.

This makes $\displaystyle A=-3$, $\displaystyle B=3$, and $\displaystyle C=-1$.

Thus you integral becomes $\displaystyle -3\int \frac{1}{x-1}\,dx + \int \frac{3x-1}{x^2-2x+10}\,dx$. (I completed the square in the denominator of the second integral.)

The first integral clearly equals $\displaystyle -3ln(x-1)$.

From now on, I'll be dealing only with the second integral.

Rewrite it like this: $\displaystyle \int \frac{3x-1-2+2}{x^2-2x+10}\,dx$.

Now, break it up into two integrals: $\displaystyle \int\frac{3x-3}{x^2-2x+10}\,dx + \int\frac{2}{x^2-2x+10}\,dx$.

In the first integral, let $\displaystyle u=x^2-2x+10$. Thus $\displaystyle du=2x-2\,dx$. Now that integral becomes $\displaystyle \frac{3}{2}\int\frac{\,du}{u} = \frac{3}{2}ln(u) = \frac{3}{2}ln(x^2+2x+10)$.

On the second integral, factor out the $\displaystyle 2$ and complete the square in the denominator to get $\displaystyle 2\int\frac{1}{(x-1)^2+9}\,dx$.

Then, factor a $\displaystyle 9$ out of the denominator to get: $\displaystyle \frac{2}{9}\int\frac{1}{\frac{(x-1)^2}{9}+1}\,dx$.

Rewrite the denominator: $\displaystyle \frac{2}{9}\int\frac{1}{(\frac{x-1}{3})^2+1}\,dx$.

Let $\displaystyle u = \frac{x-1}{3}$ and $\displaystyle du = \frac{1}{3}\,dx$.

Thus your integral becomes: $\displaystyle \frac{2}{9}\int\frac{3\,du}{u^2+1}$.

Factor out the $\displaystyle 3$ and you have $\displaystyle \frac{2}{3}\int\frac{\,du}{u^2+1} = \frac{2}{3}arctan(u) = \frac{2}{3}arctan(\frac{x-1}{3})$.

Adding all the parts together gives you: $\displaystyle -3ln(x-1) + \frac{3}{2}ln(x^2+2x+10) + \frac{2}{3}arctan(\frac{x-1}{3})$.

Thus: $\displaystyle \int\frac{2x-29}{x^3-3x^2+12x-10}\,dx = -3ln(x-1) + \frac{3}{2}ln(x^2+2x+10) + \frac{2}{3}arctan(\frac{x-1}{3})$.