Find integral of this fraction with a radical. Is my process right? I think this involves a u-sub and then partial fractions? Is this process right? Basically I learned any integral can be solved with using usub and integration by parts... and using partial fractions is just some algebraic manipulation that then relies on some definitions of integration. I just wanted to make sure my process was right before continuing?
The integral in question:
$$ \int \frac{dx}{2 \sqrt{x+3} + x}$$
So u sub first:
$ u = \sqrt{x + 3}$ so $ u^2 = x + 3$ so $ u^2 - 3 = x$ so $\frac{dx}{du} = 2u$ so $dx = 2udu$
So the integral is equivalent to:
$$ \int \frac{2udu}{2u - u^2 - 3}$$
$$= \int \frac{2udu}{u^2 + 2u - 3}$$
$$= \int \frac{2udu}{(u + 3)(u - 1)}$$
$$= \int \frac{2udu}{(u + 3)(u - 1)}$$
So from here I can use partial fractions because we have distinct linear factors in the denominator? Is that right?
So is this partial fraction formula right?
$$ \frac{2udu}{(u + 3)(u - 1)} = \frac{A}{ u + 3} + \frac{B}{u - 1}$$
so
$$ 2u = A(u - 1) + B(u + 3)$$
$$ 2u = Au - A + Bu + 3B$$
$$ 2u = (A+B)u - A + 3B$$
so
$$ 2 = A + B$$ and
$$ 0 = -A + 3B$$
$$2 = 4B$$
So $B = \frac{1}{2}$ and $A = \frac{3}{2}$
So then we got:
$$= \int \frac{2udu}{(u + 3)(u - 1)} = \int \frac{3}{2} \cdot \frac{du}{u + 3} + \frac{1}{2} \cdot \frac{du}{u-1}$$
$$= \int \frac{2udu}{(u + 3)(u - 1)} = \int \frac{3}{2} \cdot \frac{du}{u + 3} + \int \frac{1}{2} \cdot \frac{du}{u-1}$$
$$= \int \frac{2udu}{(u + 3)(u - 1)} = \frac{3}{2} \int \cdot \frac{du}{u + 3} + \frac{1}{2} \int \cdot \frac{du}{u-1}$$
So I guess formally do the right sides still need another usub? thats the only way to solve although I guess we can do that in our head eventually?
$$so \int \frac{du}{u+3}$$
if $z = u + 3$ and $dz = du$
so
$$\int \frac{dz}{z} = \ln{z} = \ln(u+3) $$
So we have:
$$= \int \frac{2udu}{(u + 3)(u - 1)} = \frac{3}{2} \ln {(u + 3)} + \frac{1}{2} \ln {(u - 1)}$$
So finally and remembering that $u = \sqrt{x+3}$
$$ \int \frac{dx}{2 \sqrt{x+3} + x} = \frac{3}{2} \ln {(\sqrt{x+3} + 3)} + \frac{1}{2} \ln {(\sqrt{x+3} - 1)} + C$$
Is that right? We get here through a u-sub, then partial fractions, then another u-sub?

Notice how $\int \frac{1}{x} dx = \ln{(abs(x))} + C$ and not $\ln{(x)} + C$, for instance, the term $\frac{1}{2}\ln{(u-1)}$ is actually $\frac{1}{2} \ln{(abs(u-1))}$. The final expression becomes:
$\int \frac{dx}{2 \sqrt{x+3} + x} = \frac{3}{2} \ln {(\sqrt{x+3} + 3)} + \frac{1}{2} \ln {(abs(\sqrt{x+3} - 1))} + C$
