Definite integral

Given the integral $I = \int_4^{6} \frac{x^2-x-4}{x^2-2x-3}\,dx$. The value of $\int_4^{6} \frac{x^2-x-4}{x^2-2x-3}\,dx$ correct to 4 decimal places is

Evaluate $\displaystyle\int_2^4 \frac{dx}{\sqrt{6x-8-x^2}}$.

The exact value of $\int_{0.2}^{2.2} xe^x\,dx$ is most nearly