I'm trying to work in my partial fractions chapter and some were easy but for whatever reason, I'm stuck now:
$$\int \frac{-3} {x2+2x+4} dx $$
What I tried:
since my denominator is of higher order and a irreducible quadratic, I set it up to start like this:
$$x-3 = \int \frac{Ax+B}{x2+2x+4} + \frac{Cx+D}{x2+2x+4} dx$$
I'm not really sure if that's right, the example in the book is a little different.
Anyway, after multiplying and collecting terms, I end up with $$x^5 (C) =0$$ $$x^4 (4C+D) =0$$ $$x^3 (A+12C+4D)=0$$ $$x^2 (2A+B+16C+12D) =0$$ $$x (4A+2B+16D+16) =1$$ $$4B+16D= -3$$
but then when I tried to solve by substitution/addition methods, my numbers don't make sense. Am I even on the right track here? We just got started in this class and I feel like I'm behind already. Thanks for any help.
You know that by substitution $y=x^2+1$ you have:
$$\int \frac{2x}{x^2+1}dx=\int \frac{1}{y}dy=\ln|y|+C$$
Also
$$\int \frac{1}{x^2+1}dx=\arctan x+C$$
You can use this two facts this way: write the dominator as $$c((\frac{x-a}{b})^2+1)$$ for some $a,b,c$ and next use substitution $x=by+a$, in this case:
$$x^2+2x+4=(x+1)^2+3=3((\frac{x+1}{\sqrt{3}})^2+1)$$.
Can you finish your task?