Question

integral (x^3 +2x^2 +3x - 2)/(x^2 +2x +2)

Answer 1

Hmm, this is very similar to your previous question. It would give a similar answer, it would include a log and an arctan.

Answer 2

I prefer here to write for you the steps that you sholud follow, and if you had a problem following the, you can ask:
1) Use long division to simplify the expression.
2) You will get this expression \(x+\frac{x-2}{x^2+2x+2}\).
The first term is clear. The coming steps are about how to integrate the second term:
1) You should split it into two terms:

Answer 3

You know that when you split it. both of the terms you get will have the same denominator. Make the first term in manner such that the top is a multiple of the derivative of the bottom.

Answer 4

It doesn't matter what number you would have in the top of the other term. You will just need to write it in the form \(\frac{a}{u^2+1}\), u can be any substitution of your choice.

Answer 5

Or I can split it for you if you want :)

Answer 6

The numbers here are much better than the ones in your previous question by the way.

Answer 7

Since you're not saying a word, I will do it for my own entertainment -.-.
\[\int\limits_{}^{}{x-2 \over x^2+2x+2}dx=\int\limits_{}^{}{x+1 \over x^2+2x+2}dx-\int\limits_{}^{}{3 \over x^2+2x+2}dx\].
For the first integral substituting \(u=x^2+2x+2 \implies du=2(x+1)dx\) yields:
\[{1 \over 2}\int\limits_{}^{}{1 \over u}du={1 \over 2}\ln |u|={1 \over 2}\ln |x^2+2x+2|.\]
For the second integral substitute \(u=x+1 \implies du=dx\):
\[-3\int\limits_{}^{}{1 \over (x+1)^2+1}=-3\int\limits {1 \over u^2+1}du=-3\tan^{-1}u=-3\tan^{-1}|x+1|\].

Answer 8

for my own entertainment!