\[1)\;\;\;\;\;\int_2^3 \frac{x^{3}+1}{x^3-x^2}dx\\by\;\;using\;\;long\;\;division =\int_2^3\frac{x^{2}+1}{x^{3}-x^{2}}+1\;dx\\=\int_2^3 \frac{x^{2}+1}{x^{3}-x^{2}}dx+\int_2^3 1dx\\ partical\;\;fraction\;\;for\;\; \frac{x^{2}+1}{x^{3}-x^{2}}\\=\frac{x^{2}+1}{x^{2}(x-1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x-1}\\=x^{2}+1=Ax(x-1)+B(x-1)+Cx^{2}\\ if\;\;x=1\\2=0+0+c\;\;\;\;\;c=2\\if\;\;x=0\\1=0+(-B)+0\;\;\;\;\;B=-1\\x^{2}+1=Ax(x-1)-(x-1)+2x^{2}\\if\;\;x=3\\10=6A-2+18\\6A=-6\;\;\;\;\;A=-1\\ Therefore\;\;we\;\;can\;\;rearrange\;\;the\;\;integral\;\;as:\\ \int_2^3 -\frac{1}{x}-\frac{1}{x^{2}}+\frac{2}{x-1}dx+ x\mid_2^3\\=-\int_2^3\frac{1}{x}dx-\int_2^3 \frac{1}{x^2}dx+2\int_2^3 \frac{1}{x-1}dx+1\\=-log(x)\mid_2^3 -(-\frac{1}{x})\mid_2^3 +2\int_2^3 \frac{1}{x-1}dx +1\\by\;\;using\;\;subsitution\;\;set\;\;u=x-1\;\;du=1dx\;\;\;x:2\;to\;3\;\;u:1\;to\;2\\=[-log(3)+log(2)]-(-\frac{1}{3}+\frac{1}{2})+2\int_1^2 \frac{1}{u}du +1\\=[-log(3)+log(2)]+\frac{5}{6}+2[log(u)]\mid_1^2\\=log(2)-log(3)+2log(2)-2log(1)+\frac{5}{6}\\=3log(2)-log(3)+\frac{5}{2}\\2)\;\;\;\;\;\int_{0}^{1} \frac{3x+11}{x^{2}-x-6}dx\\= \int_{0}^{1} \frac{3x+11}{(x+2)(x-3)}dx\\Partial\;Fraction:\;=\frac{a}{x+2} +\frac{b}{x-3} = \frac{3x+11}{(x+2)(x-3)}\\A(x-3) + B(x+2)= 3x+11\\if\;x=3 , 9+11=5B, B=4\\if\;x=n-2 , -6+11=-5A, A=-1\\so\;new\;integral\;will\;look\;like:\\ \int_{0}^{1} \frac{4}{x-3} – \frac{1}{x+2} dx\\set\;\;u=x-3 , du=dx\;\;\;\;\;and\;\;w=x+2,dw=dx\\so\;by\;using\;this\;subsituition\;our\;new\;integral\;will\;be\\=4\int_{-3}^{-2}\frac{1}{u}du – \int_{2}^{3} \frac{1}{w}dw=4log(2)-4log(3)-log(3)+log(2)=5log(2)-5log(3)\]

3) \[\int_{1}^{2} 2 \sqrt{1-e^{2x}}dx, \; \;by \;using \;using\; subsituition\; t=1-e^{2x},\;\; dt=-2e^{2x}dx, \;\;dx=\frac{-dt}{2(1-t)}, \;\; so \;\;\int_{1}^{2} 2 \sqrt{1-e^{2x}}dx=2\int_{1- e^{2}}^{1-e^{4}} \sqrt{t} \frac{-dt}{2(1-t)}= \int_{1-e^{2}}^{1-e^{4}}\frac{\sqrt{t}}{t-1}, \;using\; another\;subsituition \;u=\sqrt{t},\;\; t=u^{2}, \;\;dt=2u*du \;\;\;so \; \int_{1-e^{2}}^{1-e^{4}}\frac{\sqrt{t}}{t-1} = \;\int_{\sqrt{1-e^{2}}}^{\sqrt{1-e^{4}}} \frac{z}{z^{2}-1}*2z*dz= \;\int_{\sqrt{1-e^{2}}}^{\sqrt{1-e^{4}}} \frac{2z^{2}}{z^{2}-1}*dz=\int_{\sqrt{1-e^{2}}}^{\sqrt{1-e^{4}}} 2 +\frac{2}{z^{2}-1}*dz=\int_{\sqrt{1-e^{2}}}^{\sqrt{1-e^{4}}}2 +\frac{2}{(z-1)(z+1)}*dz, \;\;solving\; this\;\frac{2}{(z-1)(z+1)} \; by\; partial \;fraction\; we\; will \;get\; \frac{1}{z-1}\; -\frac{1}{z+1}. \int_{\sqrt{1-e^{2}}}^{\sqrt{1-e^{4}}} 2 \;
+\;\frac{2}{(z-1)(z+1)}*dz=\int_{\sqrt{1-e^{2}}}^{\sqrt{1-e^{4}}} 2 \;
+ \;\frac{1}{z-1} \;-\frac{1}{z+1}=\;\left | ln\left | z-1 \right | – ln\left | z+1 \right | + 2z \right |^{1-e^{4}}_{1-e^{2}}=\;\; ( ln(\sqrt{1-e^{4}} -1) – (ln(\sqrt{1-e^{4}} +1)) + 2\sqrt{1-e^{4}}) -\;\; ( ( ln(\sqrt{1-e^{2}} -1)\;\; – (ln(\sqrt{1-e^{2}} +1) + 2\sqrt{1-e^{2}})). This \;being \;our\;final\;answer.\]