Hence, show that integral_0¹ (2x+1)/((x-2)(x²+1)) dx = -(1)/(2) ln 8.

Let $\frac{2x+1}{(x-2)(x^2+1)} = \frac{A}{x-2} + \frac{Bx+C}{x^2+1}$.
Multiply both sides by $(x-2)(x^2+1)$:

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

To find $A$, set $x=2$:

$$2(2)+1 = A(2^2+1) + (B(2)+C)(2-2)$$ $$5 = A(4+1) + 0$$ $$5 = 5A \implies A = 1$$

Substitute $A=1$ back into the equation:

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

Compare the coefficients of powers of $x$:
For $x^2$: $0 = 1+B \implies B = -1$
For $x$: $2 = -2B+C$
Substitute $B=-1$: $2 = -2(-1)+C \implies 2 = 2+C \implies C = 0$
For the constant term: $1 = 1-2C$
Substitute $C=0$: $1 = 1-2(0) \implies 1 = 1$ (consistent)

So, the partial fraction decomposition is:

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

We need to evaluate $\int_{0}^{1} \left( \frac{1}{x-2} - \frac{x}{x^2+1} \right) dx$.

The integral of the first term is:

$$\int \frac{1}{x-2} dx = \ln|x-2|$$

For the second term, let $u = x^2+1$, so $du = 2x \, dx \implies x \, dx = \frac{1}{2} du$:

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

(Since $x^2+1$ is always positive, we can remove the absolute value.)

Combining these, the indefinite integral is:

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

$$\left[ \ln|x-2| - \frac{1}{2} \ln(x^2+1) \right]_{0}^{1}$$

Evaluate at the upper limit $x=1$:

$$\ln|1-2| - \frac{1}{2} \ln(1^2+1) = \ln|-1| - \frac{1}{2} \ln(2) = \ln(1) - \frac{1}{2} \ln(2) = 0 - \frac{1}{2} \ln(2) = -\frac{1}{2} \ln(2)$$

Evaluate at the lower limit $x=0$:

$$\ln|0-2| - \frac{1}{2} \ln(0^2+1) = \ln|-2| - \frac{1}{2} \ln(1) = \ln(2) - \frac{1}{2}(0) = \ln(2)$$

Subtract the lower limit value from the upper limit value:

$$\left( -\frac{1}{2} \ln(2) \right) - \left( \ln(2) \right) = -\frac{1}{2} \ln(2) - \frac{2}{2} \ln(2) = -\frac{3}{2} \ln(2)$$

We need to show that $-\frac{3}{2} \ln(2)$ is equal to $-\frac{1}{2} \ln 8$.
Using logarithm properties, $\ln(a^b) = b \ln a$:

$$-\frac{1}{2} \ln 8 = -\frac{1}{2} \ln (2^3) = -\frac{1}{2} \cdot 3 \ln 2 = -\frac{3}{2} \ln 2$$

Since $-\frac{3}{2} \ln 2 = -\frac{3}{2} \ln 2$, the equality holds.

Therefore, we have shown that:

$$\int_{0}^{1} \frac{2x+1}{(x-2)(x^2+1)} dx = \boxed{-\frac{1}{2} \ln 8}$$