{"id":45,"date":"2017-02-09T02:00:54","date_gmt":"2017-02-09T09:00:54","guid":{"rendered":"https:\/\/blogs.ubc.ca\/moiz12\/?p=45"},"modified":"2017-02-09T02:19:27","modified_gmt":"2017-02-09T09:19:27","slug":"45","status":"publish","type":"post","link":"https:\/\/blogs.ubc.ca\/moiz12\/2017\/02\/09\/45\/","title":{"rendered":"3 Tough Integrals to find"},"content":{"rendered":"\\[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} &#8211; \\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 &#8211; \\int_{2}^{3} \\frac{1}{w}dw=4log(2)-4log(3)-log(3)+log(2)=5log(2)-5log(3)\\]\n<p>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 \\;<br \/>\n+\\;\\frac{2}{(z-1)(z+1)}*dz=\\int_{\\sqrt{1-e^{2}}}^{\\sqrt{1-e^{4}}} 2 \\;<br \/>\n+ \\;\\frac{1}{z-1} \\;-\\frac{1}{z+1}=\\;\\left |  ln\\left | z-1 \\right | &#8211; ln\\left | z+1 \\right |  + 2z \\right |^{1-e^{4}}_{1-e^{2}}=\\;\\; ( ln(\\sqrt{1-e^{4}} -1) &#8211; (ln(\\sqrt{1-e^{4}} +1)) + 2\\sqrt{1-e^{4}}) -\\;\\; ( ( ln(\\sqrt{1-e^{2}} -1)\\;\\; &#8211; (ln(\\sqrt{1-e^{2}} +1) + 2\\sqrt{1-e^{2}})).  This \\;being \\;our\\;final\\;answer.\\]<\/p>\n","protected":false},"excerpt":{"rendered":"<p>3)<\/p>\n","protected":false},"author":45089,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-45","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/blogs.ubc.ca\/moiz12\/wp-json\/wp\/v2\/posts\/45","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blogs.ubc.ca\/moiz12\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blogs.ubc.ca\/moiz12\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blogs.ubc.ca\/moiz12\/wp-json\/wp\/v2\/users\/45089"}],"replies":[{"embeddable":true,"href":"https:\/\/blogs.ubc.ca\/moiz12\/wp-json\/wp\/v2\/comments?post=45"}],"version-history":[{"count":18,"href":"https:\/\/blogs.ubc.ca\/moiz12\/wp-json\/wp\/v2\/posts\/45\/revisions"}],"predecessor-version":[{"id":63,"href":"https:\/\/blogs.ubc.ca\/moiz12\/wp-json\/wp\/v2\/posts\/45\/revisions\/63"}],"wp:attachment":[{"href":"https:\/\/blogs.ubc.ca\/moiz12\/wp-json\/wp\/v2\/media?parent=45"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blogs.ubc.ca\/moiz12\/wp-json\/wp\/v2\/categories?post=45"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blogs.ubc.ca\/moiz12\/wp-json\/wp\/v2\/tags?post=45"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}