$(1\mathrm{.}4){\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}cos{h}^{-1}xdx$

$(1\mathrm{.}5){\displaystyle \underset{0}{\overset{\frac{3}{2}}{}}}\frac{1}{(4x^2+9{)}^{\frac{3}{2}}}dx,$ Using the substitution $x=\frac{3}{2}tanu$

$(1\mathrm{.}6),{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}sin(lnx)dx$

$(1\mathrm{.}7){\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{4x^2+3x-2}{(x+2)(x+3)(x^2+7x+12)}dx$

$(1\mathrm{.}8){\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}se{c}^4(\frac{1}{2}x)dx$

$(1\mathrm{.}9){\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{-2x+10}{5-4x-x^2}dx$

$(1\mathrm{.}10){\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}cos5x\sqrt[\phantom{2}]{1+sin5x}dx$

$(1\mathrm{.}11){\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}co{t}^{3}xcs{c}^{5}xdx$

$2$