Evaluate the Integral (Using ArcTrig)

[lude1]

Homework Statement

Evaluate the integral:

2x - 5
-------------
x^2 + 2x + 2

Homework Equations

(1/a) * arctan(u/a) + C
ln|x| (?)

There was a similar problem in the book whose answer had ln|x|, however I am not sure if it will be used for this particular problem.

The Attempt at a Solution

I started by splitting the terms up.

[2xdx / (x^2 + 2x + 2)] - [5dx / (x^2 + 2x + 2)]​

I then completed the square for the second term, resulting with this:

[(2xdx) / (x^2+2x+2)] - {5[dx/[(x+1)^2) + 1]}​

For the first section, aka: [(2xdx) / (x^2+2x+2)], this is what I have so far.

u = x^2+2x+2
du = (2x + 2)dx​

It would have been okay if it was just 2xdx (I would have ended up with ln|x^2+2x+2|), but I have a 2dx since the dx is being multiplied. I don't think I can divide the 2dx and end up with

(1/2)dudx = 2xdx​

Because I will have dudx and that doesn't make sense. Which ultimately leads me to my final conclusion - I'm doing something wrong. Maybe I have to end up with an natural log somehow? But if so, I don't know how to get it.

For the second section, aka: {5[dx/[((x+1)^2) + 1]]}, this is what I got as an answer.

u^2 = (x+1)^2
u = x+1
du = dx
a^2 = 1
a = 1​

Thus,

5[arctan(x+1)] + C​

I hope that is correct?

Now the only problem is figuring out the first section :/

 

[Gib Z]

Welcome to Physicsforums. Nice to see you are following forum guidelines and presenting us with the attempted work.

You will do a lot of these problems so it's best to learn a pattern for them: When you have the integral of a linear over a quadratic, you will usually get an arctangent term and a natural log term, with some special cases having just 1 of the two. So the best thing to do is to get the natural log term straight away by forcing the derivative into the numerator.

In this case, the denominator is x^2 + 2x + 2, so its derivative is 2x+2. Since you have 2x-5, add and subtract 7 from the numerator. So then split the numerator with one having (2x+2) and the other having (-7), as you have two very simple integrals left over.