(a) Find ∫ (x +2)/(√(x^2-4 x -5)) d x.

$\displaystyle \int \frac{\mathrm{x}+2}{\sqrt{\mathrm{x}^{2}-4 \mathrm{x}-5}} \mathrm{dx}=\frac{1}{2} \int \frac{2 \mathrm{x}-4}{\sqrt{\mathrm{x}^{2}-4 \mathrm{x}-5}} \mathrm{dx}+4 \int \frac{1}{\sqrt{(\mathrm{x}-2)^{2}-3^{2}}} \mathrm{dx}$

$$\begin{gathered} =\sqrt{\mathrm{x}^{$\displaystyle 2$}-$\displaystyle 4$ \mathrm{x}-$\displaystyle 5$}+$\displaystyle 4$ \log \left|\mathrm{x}-$\displaystyle 2$+\sqrt{\mathrm{x}^{$\displaystyle 2$}-$\displaystyle 4$ \mathrm{x}-$\displaystyle 5$}\right|+c \text { Or } \end{gathered}

$$(b) $\displaystyle \mathrm{I}=\int_{-\mathrm{a}}^{\mathrm{a}} \frac{9^{\mathrm{x}}}{1+9^{\mathrm{x}}} \mathrm{d} \mathrm{x}$

Replacing ' $\displaystyle \mathbf{x}$ ' by $\displaystyle -\mathbf{a}+\mathbf{a}-\mathbf{x}$

$$\begin{equation*} \mathrm{I}=\int_{-\mathrm{a}}^{\mathrm{a}} \frac{$\displaystyle 9$^{\mathrm{a}-\mathrm{a}-\mathrm{x}}}{$\displaystyle 1$+$\displaystyle 9$^{\mathrm{a}-\mathrm{a}-\mathrm{x}}} \mathrm{d} \mathrm{x}=\int_{-\mathrm{a}}^{\mathrm{a}} \frac{$\displaystyle 9$^{-\mathrm{x}}}{$\displaystyle 1$+$\displaystyle 9$^{-\mathrm{x}}} \mathrm{d} \mathrm{x}=\int_{-\mathrm{a}}^{\mathrm{a}} \frac{1}{$\displaystyle 1$+$\displaystyle 9$^{\mathrm{x}}} \mathrm{d} \mathrm{x} \tag{ii} \end{equation*}

$$Adding (i) & (ii) we get, $\displaystyle \left.2 \mathrm{I}=\int_{-\mathrm{a}}^{\mathrm{a}} \mathrm{dx}=\mathrm{x}\right]_{-\mathrm{a}}{ }^{\mathrm{a}}=2 \mathrm{a} \Rightarrow \mathrm{I}=\mathrm{a}$