(a) Find: ∫ (x^2+1)/(( x -1)^2( x +3)) d x

$\displaystyle \frac{\mathrm{x}^{2}+1}{(\mathrm{x}-1)^{2}(\mathrm{x}+3)}=\frac{\mathrm{A}}{\mathrm{x}-1}+\frac{\mathrm{B}}{(\mathrm{x}-1)^{2}}+\frac{\mathrm{C}}{\mathrm{x}+3}=\frac{3 / 8}{\mathrm{x}-1}+\frac{1 / 2}{(\mathrm{x}-1)^{2}}+\frac{5 / 8}{\mathrm{x}+3}$

\[\mathrm{I}=\frac{3}{8} \log |\mathrm{x}-1|-\frac{1}{2(\mathrm{x}-1)}+\frac{5}{8} \log |\mathrm{x}+3|+\mathrm{C} \]

Let $\displaystyle \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\mathrm{x}}{\sin \mathrm{x}+\cos \mathrm{x}} d \mathrm{x}$

\[\mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\dfrac{\pi}{2}-\mathrm{x}}{\cos \mathrm{x}+\sin \mathrm{x}} d \mathrm{x} \]

using property

$\displaystyle 2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\dfrac{\pi}{2}}{\sin \mathrm{x}+\cos \mathrm{x}} \mathrm{dx}$

\[=\frac{\pi}{2} \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sin \left(\dfrac{\pi}{4}+\mathrm{x}\right)} \mathrm{dx} \]

\[2 \mathrm{I}=\frac{\pi}{2 \sqrt{2}} \log \left|\operatorname{cosec}\left(\frac{\pi}{4}+\mathrm{x}\right)-\cot \left(\frac{\pi}{4}+\mathrm{x}\right)\right|_{0}^{\frac{\pi}{2}} \]