Workout the following:

$Ap-24$

a$){\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(x-3{)}^2\sqrt[\phantom{2}]{6x-x^2}dx$

b$){\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{1+sinx}{sinx(1+cosx)}dx$

c$){\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{si{n}^{-1}x}{(1-x^2{)}^{\frac{1}{2}}}dx52$

$2.$ Workout the following:

Oct$-23$

a$){\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{dx}{x^2(a+bx{)}^2}dx$

b$){\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{dx}{\sqrt[\phantom{2}]{2+3x-2x^2}}dx$

c$){\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{si{n}^{-1}x}{sinx(1+cosx)}dx$

$3.$ Workout the following:

$[$Ap$-23]$

a$){\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{x^2+4}{x^2+2x+3}dx$

b$){\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{dx}{(2x+3)\sqrt[\phantom{2}]{x^2+3x+2}}$

c$){\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{sin2xdx}{(a+bcosx{)}^2}$