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