We are asked to evaluate the integral $\int \frac{2x+1}{(x-1)(x^2+1)} dx$.

1. Problem Description

We are asked to evaluate the integral $\int \frac{2x+1}{(x-1)(x^2+1)} dx$.

2. Solution Steps

First, we decompose the integrand into partial fractions. We want to find constants A, B, and C such that:

$$\frac{2x+1}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}$$

Multiplying both sides by $(x-1)(x^2+1)$, we have:

$$2x+1 = A(x^2+1) + (Bx+C)(x-1)$$

$$2x+1 = Ax^2 + A + Bx^2 - Bx + Cx - C$$

$$2x+1 = (A+B)x^2 + (C-B)x + (A-C)$$

Equating coefficients, we get the following system of equations:

$$A+B = 0$$

$$C-B = 2$$

$$A-C = 1$$

From the first equation, $B = -A$.

Substituting into the second equation, $C+A = 2$.

Adding this to the third equation, $A-C = 1$, we get $2A = 3$, so $A = \frac{3}{2}$.

Then $B = -A = -\frac{3}{2}$.

And $C = A-1 = \frac{3}{2} - 1 = \frac{1}{2}$.

Therefore,

$$\frac{2x+1}{(x-1)(x^2+1)} = \frac{\frac{3}{2}}{x-1} + \frac{-\frac{3}{2}x + \frac{1}{2}}{x^2+1} = \frac{3}{2(x-1)} + \frac{-3x+1}{2(x^2+1)}$$

Now we can integrate:

$$\int \frac{2x+1}{(x-1)(x^2+1)} dx = \int \left( \frac{3}{2(x-1)} + \frac{-3x+1}{2(x^2+1)} \right) dx$$

$$= \frac{3}{2} \int \frac{1}{x-1} dx - \frac{3}{2} \int \frac{x}{x^2+1} dx + \frac{1}{2} \int \frac{1}{x^2+1} dx$$

The first integral is $\frac{3}{2} \ln|x-1|$.

For the second integral, we can use the substitution $u = x^2+1$, so $du = 2x dx$, and $x dx = \frac{1}{2}du$.

$$\int \frac{x}{x^2+1} dx = \int \frac{1}{2u} du = \frac{1}{2} \ln|u| = \frac{1}{2} \ln(x^2+1)$$

(Note that we don't need absolute values here since $x^2+1>0$).

So, $-\frac{3}{2} \int \frac{x}{x^2+1} dx = -\frac{3}{2} \cdot \frac{1}{2} \ln(x^2+1) = -\frac{3}{4} \ln(x^2+1)$.

The third integral is $\frac{1}{2} \int \frac{1}{x^2+1} dx = \frac{1}{2} \arctan(x)$.

Therefore,

$$\int \frac{2x+1}{(x-1)(x^2+1)} dx = \frac{3}{2} \ln|x-1| - \frac{3}{4} \ln(x^2+1) + \frac{1}{2} \arctan(x) + C$$

$$= \frac{3}{2} \ln|x-1| - \frac{1}{2} (\frac{3}{2} \ln(x^2+1)) + \frac{1}{2} \arctan(x) + C$$

Comparing with the answer options, note that $\arctan(x) = \tan^{-1}(x)$.

Also note that $\frac{3}{4} \ln(x^2+1) = \frac{1}{2} \cdot \frac{3}{2} \ln(x^2+1)$, so D looks most promising.

3. Final Answer

D. $\frac{3}{2} (ln|x-1| - \frac{1}{2} ln|x^2+1| + tan^{-1}(x)) + C$