(a) $\ln 2$

(b) (i)$-\cos (\ln x)+c$

(b)(ii) $\frac{1}{5}\left(-x^2 \cos (\ln x)+2 x^2 \sin (\ln x)\right)+c$

(a) $\int_{-3}^{-1} \frac{|x+2|}{x^2+4 x+5} \mathrm{~d} x=-\int_{-3}^{-2} \frac{x+2}{x^2+4 x+5} \mathrm{~d} x+\int_{-2}^{-1} \frac{x+2}{x^2+4 x+5} \mathrm{~d} x$
$=-\frac{1}{2} \int_{-3}^{-2} \frac{2 x+4}{x^2+4 x+5} \mathrm{~d} x+\frac{1}{2} \int_{-2}^{-1} \frac{2 x+4}{x^2+4 x+5} \mathrm{~d} x$
$=\frac{1}{2}\left(-\left[\ln \left|x^2+4 x+5\right|\right]_{-3}^{-2}+\left[\ln \left|x^2+4 x+5\right|\right]_{-2}^{-1}\right)$
$=\frac{1}{2}[(-\ln 1+\ln 2)+(\ln 2-\ln 1)]=\ln 2$

(b)(i) $\int \frac{\sin (\ln x)}{x} \mathrm{~d} x=\int \frac{1}{x} \sin (\ln x) \mathrm{d} x=-\cos (\ln x)+c$

(b)(ii) $\int x \sin (\ln x) \mathrm{d} x=\int x\left(\frac{x}{x}\right) \sin (\ln x) \mathrm{d} x$
$=\int x^2\left(\frac{1}{x}\right) \sin (\ln x) \mathrm{d} x$
$=x^2(-\cos (\ln x))-\int 2 x(-\cos (\ln x) \mathrm{d} x$
$=-x^2 \cos (\ln x)+2 \int x\left(\frac{x}{x}\right) \cos (\ln x) \mathrm{d} x$
$=-x^2 \cos (\ln x)+2\left[x^2 \sin (\ln x)-\int 2 x \sin (\ln x) \mathrm{d} x\right]$
$5 \int x \sin (\ln x) \mathrm{d} x=-x^2 \cos (\ln x)+2 x^2 \sin (\ln x)$
$\int x \sin (\ln x) \mathrm{d} x=\frac{1}{5}\left(-x^2 \cos (\ln x)+2 x^2 \sin (\ln x)\right)+c$