Partial fraction integration problem I'm trying to solve this integral by partial fraction:
$$
 \int \frac{2x-6} {(x-2)^2(x^2+4)} dx \
$$
i think i have to write the expression like
$$
 2\int \frac{x-3} {(x-2)^3(x+2)} dx \
$$
Then i don't know how i should resolve the partial fraction!
 A: Careful!

$$
 \int \frac{2x-6} {(x-2)^2(x^2+4)} dx \
$$
i think i have to write the expression like
$$
 2\int \frac{x-3} {(x-2)^3(x+2)} dx \
$$

You seem to have replaced $x^2+4$ by $(x-2)(x+2)$, but these are not the same! You might be confusing with $x^2\color{red}{-}4=(x-2)(x+2)$...
You can take out the (constant) factor $2$, but you don't have to. For the partial fraction decomposition, you're looking for numbers $A$, $B$, $C$ and $D$ such that:
$$\frac{2x-6} {(x-2)^2(x^2+4)} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{Cx+D}{x^2+4} $$
Once you have that, the integral splits into these three terms and these are all easy to integrate.
I can elaborate on the partial fraction decomposition; unless you can take it from here?

For the partial fraction decomposition:
$$\begin{array}{rl}
\displaystyle \frac{2x-6} {(x-2)^2(x^2+4)}
 & \displaystyle = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{Cx+D}{x^2+4} \\[7pt]
 & \displaystyle = \frac{A(x-2)(x^2+4)+B(x^2+4)+(Cx+D)(x-2)^2}{(x-2)^2(x^2+4)} \end{array}$$
Now equating the numerators:
$$2x-6 = A(x-2)(x^2+4)+B(x^2+4)+(Cx+D)(x-2)^2$$
Method 1
You can expand the RHS and group per power of $x$ and then identify the corresponding coefficients left and right; this gives you four linear equations in the 4 unknown variables.
$$2x-6 = (A+C)x^3+(-2A+B-4C+D)x^2+(4A+4C-4D)x+(-8A+4B+4D)$$
Which gives the system:
$$\left\{\begin{array}{rcl}
A+C & = &  0  \\
-2A+B-4C+D & = &  0  \\ 
4A+4C-4D & =&  2  \\ 
-8A+4B+4D & = & -6
\end{array} \right. \quad \Rightarrow  \quad
\left\{\begin{array}{rcl}
A & = &  \tfrac{3}{8}  \\
B & = &  -\tfrac{1}{4} \\ 
C & =&  -\tfrac{3}{8}  \\ 
D & = & -\tfrac{1}{2}
\end{array} \right.$$
Method 2
It is indeed also possible to simplify by choosing handy values for $x$ to substitute into
$$2x-6 = A(x-2)(x^2+4)+B(x^2+4)+(Cx+D)(x-2)^2$$
For example:


*

*Substitution of $x=2$ yields $-2=8B$ so $B = -\tfrac{1}{4}$.

*Substitution of $x=0$ yields $-6=-8A+4B+4D$ so $8A-4D=5$.

*(...)


If you know complex numbers, substitution of $x = 2i$ gives you $C$ and $D$ immediately, then $A$ follows from $8A-4D=5$.
