{"id":222257,"date":"2025-05-31T07:29:13","date_gmt":"2025-05-31T07:29:13","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=222257"},"modified":"2025-05-31T07:29:15","modified_gmt":"2025-05-31T07:29:15","slug":"draw-the-lewis-structure-and-3d-shape-structure-with-perspective-for-ibr4","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/31\/draw-the-lewis-structure-and-3d-shape-structure-with-perspective-for-ibr4\/","title":{"rendered":"Draw the Lewis structure and 3D shape\/structure (with perspective) for IBr4*"},"content":{"rendered":"\n

Draw the Lewis structure and 3D shape\/structure (with perspective) for IBr4*. State the electron pair geometry and the molecular geometry for your structure in #9. What is\/are the bond angle(s) in your structure in #9?<\/p>\n\n\n\n

Make a substitution to express the integrand as a rational function and then evaluate the integral. (Use C for the constant of integration.)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

We are given the integral: \u222b2e2xe2x+16ex+63\u2009dx\\int \\frac{2e^{2x}}{e^{2x} + 16e^x + 63} \\, dx<\/p>\n\n\n\n

\n\n\n\n

Step 1: Substitution<\/h3>\n\n\n\n

Let us simplify the expression by substituting: u=ex\u21d2du=exdx\u21d2dx=duuu = e^x \\Rightarrow du = e^x dx \\Rightarrow dx = \\frac{du}{u}<\/p>\n\n\n\n

Then: e2x=(ex)2=u2e^{2x} = (e^x)^2 = u^2<\/p>\n\n\n\n

Now rewrite the integral: \u222b2e2xe2x+16ex+63\u2009dx=\u222b2u2u2+16u+63\u22c51u\u2009du=\u222b2uu2+16u+63\u2009du\\int \\frac{2e^{2x}}{e^{2x} + 16e^x + 63} \\, dx = \\int \\frac{2u^2}{u^2 + 16u + 63} \\cdot \\frac{1}{u} \\, du = \\int \\frac{2u}{u^2 + 16u + 63} \\, du<\/p>\n\n\n\n

\n\n\n\n

Step 2: Use Substitution for Rational Function<\/h3>\n\n\n\n

We will now use a direct substitution to integrate the rational function:<\/p>\n\n\n\n

Let the denominator be: u2+16u+63u^2 + 16u + 63<\/p>\n\n\n\n

We complete the square: u2+16u+63=(u+8)2\u22121u^2 + 16u + 63 = (u + 8)^2 – 1<\/p>\n\n\n\n

But since this is a simple rational function, we can integrate using u-substitution<\/strong> directly.<\/p>\n\n\n\n

Let: v=u2+16u+63\u21d2dvdu=2u\u21d2du=dv2uv = u^2 + 16u + 63 \\Rightarrow \\frac{dv}{du} = 2u \\Rightarrow du = \\frac{dv}{2u}<\/p>\n\n\n\n

Then: \u222b2uv\u2009du=\u222b2uv\u22c5dv2u=\u222bdvv=ln\u2061\u2223v\u2223+C=ln\u2061\u2223u2+16u+63\u2223+C\\int \\frac{2u}{v} \\, du = \\int \\frac{2u}{v} \\cdot \\frac{dv}{2u} = \\int \\frac{dv}{v} = \\ln|v| + C = \\ln|u^2 + 16u + 63| + C<\/p>\n\n\n\n

Now recall that u=exu = e^x, so: u2+16u+63=e2x+16ex+63u^2 + 16u + 63 = e^{2x} + 16e^x + 63<\/p>\n\n\n\n

\n\n\n\n

\u2705 Final Answer:<\/h3>\n\n\n\n

\u222b2e2xe2x+16ex+63\u2009dx=ln\u2061\u2223e2x+16ex+63\u2223+C\\boxed{\\int \\frac{2e^{2x}}{e^{2x} + 16e^x + 63} \\, dx = \\ln|e^{2x} + 16e^x + 63| + C}<\/p>\n\n\n\n

\n\n\n\n

\ud83d\udcd8 Explanation <\/h3>\n\n\n\n

This integral involves an exponential expression in both the numerator and denominator. To simplify, we use substitution. Setting u=exu = e^x converts the exponential terms into polynomials, allowing us to express the integrand as a rational function. This substitution gives e2x=u2e^{2x} = u^2 and dx=duudx = \\frac{du}{u}. Substituting into the integral: 2e2xe2x+16ex+63dx=2u2u2+16u+63\u22c51u\u2009du=2uu2+16u+63\u2009du\\frac{2e^{2x}}{e^{2x} + 16e^x + 63} dx = \\frac{2u^2}{u^2 + 16u + 63} \\cdot \\frac{1}{u} \\, du = \\frac{2u}{u^2 + 16u + 63} \\, du<\/p>\n\n\n\n

This is now a rational function where the numerator is the derivative (up to a constant) of the denominator. Recognizing this structure allows us to apply a substitution v=u2+16u+63v = u^2 + 16u + 63. Then, dv=(2u+16)dudv = (2u + 16)du, but since we only have 2u\u2009du2u\\,du, we simply do: dv=2u\u2009du\u21d2dv2u=dudv = 2u \\, du \\Rightarrow \\frac{dv}{2u} = du<\/p>\n\n\n\n

Plugging in, the integral becomes: \u222b2uv\u22c5du=\u222bdvv=ln\u2061\u2223v\u2223+C\\int \\frac{2u}{v} \\cdot du = \\int \\frac{dv}{v} = \\ln|v| + C<\/p>\n\n\n\n

Back-substituting v=u2+16u+63v = u^2 + 16u + 63, and u=exu = e^x, we get the final answer: ln\u2061\u2223e2x+16ex+63\u2223+C\\ln|e^{2x} + 16e^x + 63| + C<\/p>\n\n\n\n

This method shows the power of substitution in simplifying complex-looking exponential integrals.<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

Draw the Lewis structure and 3D shape\/structure (with perspective) for IBr4*. State the electron pair geometry and the molecular geometry for your structure in #9. What is\/are the bond angle(s) in your structure in #9? Make a substitution to express the integrand as a rational function and then evaluate the integral. (Use C for the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-222257","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222257","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=222257"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222257\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=222257"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=222257"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=222257"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}