Indefinite Integral of $(x^2+1)\over (x^3 + 8)$ I need help in solving this indefinite integral: $$\int {(x^2+1)\over (x^3 + 8)} $$ 
I know it needs to be reduced to the form $A\over (x+2)$ + $B\over (x+2)^2$ where A and B are constants, but I cannot seem to solve this integral. 
Thanks in advance for any help.

$$(x^3 + 8) = (x + 2)(x^2 - 2x + 4) \neq (x+2)^3$$
To break the integrand into partial fractions, use $$\int {x^2+1\over x^3 + 8}\,dx = \int \left(\frac{A}{x+2}+\frac{Bx+C}{x^2-2x+4}\right)\,dx$$
The first term will integrate as $\;A\log|x + 2| + \text{constant}$. 
Once you solve for $A, B, C$, you'll get a better idea how to integrate the last term: whether the numerator becomes the derivative of the denominator (if $B = 2$ and $C = -2$), or if you need to split it into the sum of two terms and use some other approach, for example, completing the square to get $\;x^2 - 2x + 4 = (x - 1)^2 + 3\;$ and using trig substitution if/when needed.

Let's pin down our constants $A, B, C$:
We need for $$A(x^2 - 2x + 4) + (Bx + C)(x + 2) = x^2 + 1$$
Setting $x = -2$, we can solve for $A$: $$A(4+4+4) + (Bx + C)\cdot (0) = 4 + 1 \implies 12A= 5 \iff A = \dfrac 5{12}.$$
Now, we can expand the left-hand side, equate coefficients, and solve for $B, C$ knowing $A$:
$$\begin{align} Ax^2 - 2Ax + 4A + Bx^2 + (2B + C) x + 2C & = (A + B)x^2 + (-2A + 2B + C)x + (4A + 2C) \\ \\ & = 1\cdot x^2 + 0\cdot x + 1 \iff \end{align}$$ 
$$ (A+B) = 1 \iff B = 1 - A = 1 - \dfrac 5{12} = \dfrac 7{12}$$
$$4A + 2C = 1 \iff C = \frac 12(1 - 4A) = \frac 12\left(1 - 4\cdot \frac 5{12}\right) = -\frac 13$$
