# How do I go about integrating this?

1. Jan 13, 2010

### JFonseka

1. The problem statement, all variables and given/known data

Find $$\int$$$$\frac{x+2}{x^{2}+2*x+10}$$*dx

You are given that $$\int$$$$\frac{du}{u^{2}+a^{2}}$$ = $$\frac{1}{a}$$tan$$^{-1}$$ $$\frac{u}{a}$$ + C for a not equal to 0

2. Relevant equations

None.

3. The attempt at a solution

I'm not entirely too sure how to go about doing this. I first thought of integration by parts, but given that this is a 3 mark question it seems quite long winded.

I split up the expression:

1/x^2+2*x+10*x+2*dx

Then I set u = x+2, du = 1, dv =1/x^2+2*x+10, v = (1/3)*arctan((1/3)*x+1/3)

I have no idea how to get v, that was calculated using Maple, so how is that arctan component calculated by hand?

Using the uv - int(vdu) formula for the rest of it results in some bizarre answers, which might be right but seems altogether too long and complex for a 3 mark answer.

The final answer as calculated by Maple is:

(1/2)*ln(x^2+2*x+10)+(1/3)*arctan((1/3)*x+1/3)

Any help and direction greatly appreciated.

2. Jan 13, 2010

### HallsofIvy

Staff Emeritus
Please use parentheses to show what you mean! I finally realized that this was (1/(x^2+ 2x+ 10))(x+2). What you wrote would be more readily interpreted as 1/(x^2+ 2x+ 10x+ 2).

Don't do this "by parts", complete the square in the denominator. $x^2+ 2x+ 10= x^2+ 2x+ 1- 1+10= (x+1)^2+ 9$. Now you can let u= x+1 so that du= dx and x= u- 1. The integral becomes
$$\int \frac{u- 1}{u^2+ 9} du= \int \frac{u}{u^2+9} du- \int \frac{1}{u^2+9}du$$
The first integral can be done with the substitution $v= u^2+9$ and 1/(u^2+9) is exactly what you need for your arctan integral.

3. Jan 13, 2010

### JFonseka

HallsofIvy, thank you so much for your response, that was a lot of help!

Sorry about the lack of parentheses.