# Rational Function Integration

• Mar 2nd 2009, 09:20 AM
My Little Pony
Rational Function Integration
Find the indefinite integral of the function:
• Mar 2nd 2009, 09:29 AM
Jameson
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.
• Mar 2nd 2009, 10:15 AM
redsoxfan325

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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