Help Integrate a problem?

First of all, we can factor a 2 out of the numerator. This gives us:

2 * integral: ( x^2 + 2x + 11 ) / ( x^2 + 2x + 10 ) dx

Now as you said, we divide the terms inside the integral. This now makes it:

2 * integral: [ 1 + ( 1 / x^2 + 2x + 10 ) ] dx

Certainly the 1 will be easy to integrate, so we focus on the other part of this. Notice we have a constant on top divided by a quadratic polynomial. We cannot use a substitution because we'd need a linear term on top if we let u equal the denominator. So, looking back at other integration techniques, it turns out this is an arctangent integration. Here's the formula for the integral of this type of expression:

(1/a)*arctan(u/a) = integral [ du / ( u^2 + a^2 ) ]

Now the trick will be getting our expression into that form. On the denominator we have x^2 + 2x + 10. What we'll end up with here is a binomial squared (which becomes u in the above formula) and a constant squared (which becomes a in the above formula). This requires that we complete the square using the first two terms of that trinomial. We take the x^2 + 2x. What third term completes that square? Remember the pattern is that we take the coefficient of x, divide by 2, and square the result. This means we'll need a +1 at the end to complete the square. So we take the +10 that we have and split it up into two pieces: +1 and +9. Now we write our denominator:

x^2 + 2x + 1 + 9

And by factoring the first three terms, we have:

(x+1)^2 + 9

Now our integral is:

2 * integral: [ 1 + ( 1 / (x+1)^2 + 9 ) ] dx

The integral of 1 is of course x. For the integral of the other part, notice that u = x+1 and a=3. Then it becomes: (1/3)arctan[(x+1)/3]. Don't forget our 2 that we had outside. Multiplying each of these by 2, our final answer is:

2x + (2/3)arctan[(x+1)/3] + C

Note: Notice that we cannot use partial fractions in this case because the denominator is not factorable.