 In this video, we'll work through another example applying the method of partial fractions to integrate a rational function, this time with a quadratic factor that is not factorable. Consider the integral of x squared plus x minus 2 over 3x minus 1 times x squared plus 1. Our goal is to decompose the integrand into partial fractions so that we can use known methods of integration to evaluate the integral. Let's focus on the integrand x squared plus x minus 2 over 3x minus 1 times x squared plus 1. As we've seen with previous examples, we can rewrite this, the integrand, as a sum of rational functions. 3x minus 1 is a linear factor, so I have a over 3x minus 1 plus. Now, x squared plus 1 is a quadratic factor that is irreducible. In other words, I cannot decompose it into real number factored form. In that case, my numerator is bx plus c to account for an x term and a potential constant term. If I had a cubic factor in the denominator that was not factorable further, then my numerator would be a standard form of a quadratic term. So, again, the numerator in the part of the decomposition with a quadratic factor as the denominator is a linear function as opposed to a constant over a linear factor. Our goal as before is to solve for the constants a, b, and c. We multiply both sides of this equation by the original denominator again since it's the common denominator among all rational expressions. This then gives us x squared plus x minus 2 equals a times x squared plus 1 plus bx plus c times 3x minus 1. We see that here again. I'm going to expand by distributing a and expanding this expression bx plus c times 3x minus bx plus 3cx minus c. I'm going to clean this up a little bit. I get ax squared plus a plus 3bx squared minus bx plus 3cx minus c. I'm going to collect like terms here. I have a plus 3b times x squared plus negative b plus 3cx plus a minus c. So we have this expression rewritten here. Again, as the equals sign implies, the left side is equivalent to the right side. I can actually equate the coefficients of each of these terms to help me solve for a, b, and c. So equating coefficients, we get one equation which is a plus 3b is equal to the coefficient of 1. We see that negative b plus 3c is also equal to 1 and a minus c is equal to negative 2. So we have three equations and three unknowns. So we're going to use processes used to solve systems of equations to find a, b, and c. So here are three equations. If I subtract equation 3 from equation 1, I see that I get 3b plus c, 1 minus a negative 2 equals 3. Using equation 2 and this new equation, I can use the method of elimination by multiplying the second equation here, this is equation 2, by 3, which gives me negative 3b plus 9c equals 3. If I add these two, I get 10c equals 6, so c is 3 fifths. Knowing this, using equation 3, a minus 3 fifths equals negative 2, which then gives me that a is negative 7 fifths. Using then equation 1, we find that negative 7 fifths plus 3b equals 1, so 3b is equal to 12 fifths, which then means that b is equal to 4 fifths. So we found our three constants, so we have the following. Negative 7 fifths, that was our a, over 3x minus 1, plus 4 fifths, x, this was our b, plus 3 fifths, which was our c, over x squared plus 1. I'm going to factor out a 1 fifth, because that's common among all terms. This leaves me with the following. And when I integrate this, I get negative 7 fifths, times the integral of 1 over 3x minus 1, plus 4 fifths, times the integral of x over x squared plus 1, plus 3 fifths, times the integral of 1 over x squared plus 1. And now we can evaluate. Negative 7 fifths, times 1 third, times the natural log of absolute value of 3x minus 1, plus 2 fifths, times the natural log of absolute value of x squared plus 1. You may want to verify this one using u substitution, plus 3 fifths, times inverse tangent of x, plus our constant c. This ultimately then gives us negative 7 over 15, times the natural log of 3x minus 1, plus 2 fifths, times natural log of x squared plus 1, plus 3 fifths, times inverse tangent of x, plus c.