First, split the fraction:

\(\frac{x(x+1)}{x^2+4} = 1 + \frac{x-4}{x^2+4}\)

Integrate the first part:

\(\int 1 \, dx = x\)

For the second part, use partial fraction decomposition:

\(\int \frac{x-4}{x^2+4} \, dx = \frac{1}{2} \ln(x^2+4) - 2 \arctan\left(\frac{x}{2}\right)\)

Evaluate from 0 to 6:

\(\left[ x + \frac{1}{2} \ln(x^2+4) - 2 \arctan\left(\frac{x}{2}\right) \right]_{0}^{6}\)

Calculate at the bounds:

At \(x = 6\):

\(6 + \frac{1}{2} \ln(36+4) - 2 \arctan 3\)

At \(x = 0\):

\(0 + \frac{1}{2} \ln 4 - 2 \arctan 0\)

Subtract the results:

\(\left( 6 + \frac{1}{2} \ln 40 - 2 \arctan 3 \right) - \left( \frac{1}{2} \ln 4 \right)\)

Simplify:

\(6 + \frac{1}{2} \ln 10 - 2 \arctan 3\)