{"id":8930,"date":"2024-01-28T19:15:56","date_gmt":"2024-01-28T13:45:56","guid":{"rendered":"https:\/\/physicscatalyst.com\/article\/?p=8930"},"modified":"2024-01-28T19:16:35","modified_gmt":"2024-01-28T13:46:35","slug":"integration-of-1-a2-x2","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/integration-of-1-a2-x2\/","title":{"rendered":"integration of 1\/a^2+ x^2, 1\/a^2- x^2, 1\/x^2 -a^2"},"content":{"rendered":"\n

Integration of 1\/a^2+ x^2, 1\/a^2- x^2, 1\/x^2 -a^2 can be obtained using trigonometric substitution, integration by partial fractions. The formula given are<\/p>\n\n\n\n

$\\int \\frac {1}{x^2 + a^2} dx = \\frac {1}{a} \\tan ^{-1} (\\frac {x}{a}) + C$<\/p>\n\n\n\n

$\\int \\frac {1}{x^2 – a^2} dx = \\frac {1}{2a} ln |\\frac {x-a}{x+a}| + C$<\/p>\n\n\n\n

$\\int \\frac {1}{a^2 – x^2} dx = \\frac {1}{2a} ln |\\frac {a+x}{a-x}| + C$<\/p>\n\n\n\n

Proof of integration of 1\/a^2+ x^2<\/h2>\n\n\n\n

This can be derived using trigonometric substitution and trigonometric identities<\/p>\n\n\n\n

$\\int \\frac {1}{x^2 + a^2} dx = \\frac {1}{a} \\tan ^{-1} (\\frac {x}{a}) + C$
Proof<\/strong>
Put $x =a \\tan \\theta$
then
$dx= a \\sec^2 \\theta d\\theta$
Therefore
$\\int \\frac {1}{x^2 + a^2} dx$
$=\\int \\frac {a \\sec^2 \\theta}{a^2 \\tan^2 \\theta + a^2} d\\theta$
Now $\\sec^2 \\theta = \\tan^2 \\theta + 1$<\/p>\n\n\n\n

Therefore
$=\\frac {1}{a} \\int d\\theta= \\frac {1}{a} \\theta + C $<\/p>\n\n\n\n

Substituting back the values<\/p>\n\n\n\n

$= \\frac {1}{a} \\tan ^{-1} (\\frac {x}{a}) + C$<\/p>\n\n\n\n

Proof of integration of 1\/a^2 – x^2<\/h2>\n\n\n\n

This can be derived using integration by partial fractions<\/p>\n\n\n\n

$\\int \\frac {1}{a^2 – x^2} dx = \\frac {1}{2a} ln |\\frac {a+x}{a-x}| + C$
Proof<\/strong>
Using partial fraction technique
$\\frac {1}{a^2 – x^2} = \\frac {A}{a-x} + \\frac {B}{a+x}$<\/p>\n\n\n\n

Solving for A and B , we get <\/p>\n\n\n\n

$\\frac {1}{a^2 – x^2} =\\frac {1}{2a}[ \\frac {1}{a-x} + \\frac {1}{a+x}]$
So
$\\int \\frac {1}{a^2 – x^2} dx $
$=\\frac {1}{2a}[ \\int \\frac {1}{a-x} dx + \\int \\frac {1}{x+a}]$
$= \\frac {1}{2a}[-ln |a-x| + ln |a+x| + C$
$=\\frac {1}{2a} ln |\\frac {a+x}{a-x}| + C$<\/p>\n\n\n\n

Proof of integration of 1\/x^2 – a^2<\/h2>\n\n\n\n

This can be derived using integration by partial fractions<\/p>\n\n\n\n

$\\int \\frac {1}{x^2 – a^2} dx = \\frac {1}{2a} ln |\\frac {x-a}{x+a}| + C$
Proof<\/strong><\/p>\n\n\n\n

Using partial fraction technique
$\\frac {1}{x^2 – a^2} = \\frac {A}{x-a} + \\frac {B}{x+a}$<\/p>\n\n\n\n

Solving for A and B , we get <\/p>\n\n\n\n

$\\frac {1}{x^2 – a^2} =\\frac {1}{2a}[ \\frac {1}{x-a} – \\frac {1}{x+a}]$
So
$\\int \\frac {1}{x^2 – a^2} dx $
$=\\frac {1}{2a}[ \\int \\frac {1}{x-a} dx – \\int \\frac {1}{x+a}]$
$= \\frac {1}{2a}[ln |x-a| – ln |x+a| + C$
$=\\frac {1}{2a} ln |\\frac {x-a}{x+a}| + C$<\/p>\n\n\n\n

Integration Based on the above<\/h2>\n\n\n\n

$\\int \\frac {1}{ax^2 + bx + c} dx$

This integral can be converted to the form A,B, C given above using completing the square method and can be evaluated using the formula<\/p>\n\n\n\n

Solved Examples<\/h2>\n\n\n\n

Question 1:<\/strong><\/p>\n\n\n\n

$$ \\int \\frac{1}{x^2 + 4} \\, dx $$<\/p>\n\n\n\n

Solution<\/strong><\/p>\n\n\n\n

In this case, $ a^2 = 4 $, so $ a = 2 $. Using the formula:<\/p>\n\n\n\n

$$ \\int \\frac {1}{x^2 + a^2} \\, dx = \\frac {1}{a} \\tan ^{-1} \\left( \\frac {x}{a} \\right) + C $$<\/p>\n\n\n\n

we get:<\/p>\n\n\n\n

$$ \\int \\frac{1}{x^2 + 4} \\, dx = \\frac {1}{2} \\tan ^{-1} \\left( \\frac {x}{2} \\right) + C $$<\/p>\n\n\n\n

Question 2:<\/strong><\/p>\n\n\n\n

$$ \\int \\frac{1}{9x^2 + 1} \\, dx $$<\/p>\n\n\n\n

Solution<\/strong><\/p>\n\n\n\n

Here, we need to bring the denominator to the form $ x^2 + a^2 $. We can rewrite $ 9x^2 + 1 $ as $ (3x)^2 + 1^2 $. Thus, $ a = 1 $ and the integral becomes:<\/p>\n\n\n\n

$$ \\int \\frac{1}{9x^2 + 1} \\, dx = \\int \\frac{1}{(3x)^2 + 1^2} \\, dx $$<\/p>\n\n\n\n

Using the formula:<\/p>\n\n\n\n

$$ \\int \\frac {1}{x^2 + a^2} \\, dx = \\frac {1}{a} \\tan ^{-1} \\left( \\frac {x}{a} \\right) + C $$<\/p>\n\n\n\n

we get:<\/p>\n\n\n\n

$$ \\int \\frac{1}{9x^2 + 1} \\, dx = \\frac {1}{3} \\tan ^{-1} (3x) + C $$<\/p>\n\n\n\n

Question 3:<\/strong><\/p>\n\n\n\n

$$ \\int \\frac{1}{9 – x^2} \\, dx $$<\/p>\n\n\n\n

Solution<\/strong><\/p>\n\n\n\n

In this case, $ a^2 = 9 $, so $ a = 3 $. Using the formula, the integral becomes:<\/p>\n\n\n\n

$$ \\int \\frac{1}{9 – x^2} \\, dx = \\frac{1}{2 \\times 3} \\ln \\left| \\frac{3 + x}{3 – x} \\right| + C $$
$$ = \\frac{1}{6} \\ln \\left| \\frac{3 + x}{3 – x} \\right| + C $$<\/p>\n\n\n\n

Question 4:<\/strong><\/p>\n\n\n\n

$$ \\int \\frac{1}{1 – 4x^2} \\, dx $$
Solution<\/strong><\/p>\n\n\n\n

Here, we can rewrite $ 1 – 4x^2 $ as $ (1)^2 – (2x)^2 $. Thus, $ a = 1 $ and the integral becomes:<\/p>\n\n\n\n

$$ \\int \\frac{1}{1 – 4x^2} \\, dx = \\int \\frac{1}{1^2 – (2x)^2} \\, dx $$<\/p>\n\n\n\n

Using the formula, we get:<\/p>\n\n\n\n

$$ \\int \\frac{1}{1 – 4x^2} \\, dx = \\frac{1}{2} \\ln \\left| \\frac{1 + 2x}{1 – 2x} \\right| + C $$<\/p>\n\n\n\n

Question 5<\/strong><\/p>\n\n\n\n

$$\\int \\frac {1}{x^2 + 2x + 7} \\; dx$$<\/p>\n\n\n\n

Solution<\/strong><\/p>\n\n\n\n

To integrate the function $\\int \\frac {1}{x^2 + 2x + 7} \\, dx$, we can use a method involving completing the square and then applying a standard integral formula. The goal is to transform the quadratic in the denominator into a form that resembles the sum of a square and a constant.<\/p>\n\n\n\n

Complete the Square:
The complete square form of $x^2 + 2x + 7$ is $(x + 1)^2 + 6$.<\/p>\n\n\n\n

Apply the Integral Formula:
The integral now takes the form $\\int \\frac {1}{(x + 1)^2 + 6} \\, dx$. This resembles the integral of the form $\\int \\frac {1}{u^2 + a^2} \\, du$, which is solved as $\\frac {1}{a} \\tan ^{-1} \\left( \\frac {u}{a} \\right) + C$.<\/p>\n\n\n\n

Let $u = x + 1$ and $a^2 = 6$ (so $a = \\sqrt{6}$), then du =dx<\/p>\n\n\n\n

$$ \\int \\frac {1}{x^2 + 2x + 7} \\, dx = \\int \\frac {1}{u^2 + (\\sqrt{6})^2} \\, du $$
Applying the formula
$= \\frac{1}{\\sqrt{6}} \\tan^{-1}\\left(\\frac{u}{\\sqrt{6}(}\\right) + C$
Substituting back
$=\\frac{1}{\\sqrt{6}} \\tan^{-1}\\left(\\frac{\\sqrt{6}(x + 1)}{6}\\right) + C $$<\/p>\n\n\n\n

\n

Other Integration Related Articles<\/p>\n\n\n\n

Integration of modulus function<\/a><\/td>Integration of tan x<\/a><\/td><\/td><\/td><\/tr>
Integration of sec x<\/a><\/td>Integration of greatest integer function<\/a><\/td><\/td><\/td><\/tr>
Integration of trigonometric functions<\/a><\/td>Integration of cot x<\/a><\/td><\/td><\/td><\/tr>
Integration of cosec x<\/a><\/td>Integration of sinx<\/a><\/td><\/td><\/td><\/tr>
Integration of irrational functions<\/a><\/td>integration of root x<\/a><\/td><\/td><\/td><\/tr>
Integration of fractional part of x<\/a><\/td>Integration of cos square x<\/a><\/td><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"

Integration of 1\/a^2+ x^2, 1\/a^2- x^2, 1\/x^2 -a^2 can be obtained using trigonometric substitution, integration by partial fractions. The formula given are $\\int \\frac {1}{x^2 + a^2} dx = \\frac {1}{a} \\tan ^{-1} (\\frac {x}{a}) + C$ $\\int \\frac {1}{x^2 – a^2} dx = \\frac {1}{2a} ln |\\frac {x-a}{x+a}| + C$ $\\int \\frac {1}{a^2 – […]<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","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":"","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":"set","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":[498],"tags":[],"class_list":["post-8930","post","type-post","status-publish","format-standard","hentry","category-maths"],"yoast_head":"\nintegration of 1\/a^2+ x^2, 1\/a^2- x^2, 1\/x^2 -a^2 - physicscatalyst's Blog<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/physicscatalyst.com\/article\/integration-of-1-a2-x2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"integration of 1\/a^2+ x^2, 1\/a^2- x^2, 1\/x^2 -a^2 - physicscatalyst's Blog\" \/>\n<meta property=\"og:description\" content=\"Integration of 1\/a^2+ x^2, 1\/a^2- x^2, 1\/x^2 -a^2 can be obtained using trigonometric substitution, integration by partial fractions. The formula given are $int frac {1}{x^2 + a^2} dx = frac {1}{a} tan ^{-1} (frac {x}{a}) + C$ $int frac {1}{x^2 – a^2} dx = frac {1}{2a} ln |frac {x-a}{x+a}| + C$ $int frac {1}{a^2 – […]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/physicscatalyst.com\/article\/integration-of-1-a2-x2\/\" \/>\n<meta property=\"og:site_name\" content=\"physicscatalyst's Blog\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/PhysicsCatalyst\" \/>\n<meta property=\"article:author\" content=\"https:\/\/www.facebook.com\/PhysicsCatalyst\" \/>\n<meta property=\"article:published_time\" content=\"2024-01-28T13:45:56+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2024-01-28T13:46:35+00:00\" \/>\n<meta name=\"author\" content=\"physicscatalyst\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"physicscatalyst\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"4 minutes\" \/>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"integration of 1\/a^2+ x^2, 1\/a^2- x^2, 1\/x^2 -a^2 - physicscatalyst's Blog","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/physicscatalyst.com\/article\/integration-of-1-a2-x2\/","og_locale":"en_US","og_type":"article","og_title":"integration of 1\/a^2+ x^2, 1\/a^2- x^2, 1\/x^2 -a^2 - physicscatalyst's Blog","og_description":"Integration of 1\/a^2+ x^2, 1\/a^2- x^2, 1\/x^2 -a^2 can be obtained using trigonometric substitution, integration by partial fractions. The formula given are $int frac {1}{x^2 + a^2} dx = frac {1}{a} tan ^{-1} (frac {x}{a}) + C$ $int frac {1}{x^2 – a^2} dx = frac {1}{2a} ln |frac {x-a}{x+a}| + C$ $int frac {1}{a^2 – […]","og_url":"https:\/\/physicscatalyst.com\/article\/integration-of-1-a2-x2\/","og_site_name":"physicscatalyst's Blog","article_publisher":"https:\/\/www.facebook.com\/PhysicsCatalyst","article_author":"https:\/\/www.facebook.com\/PhysicsCatalyst","article_published_time":"2024-01-28T13:45:56+00:00","article_modified_time":"2024-01-28T13:46:35+00:00","author":"physicscatalyst","twitter_misc":{"Written by":"physicscatalyst","Est. reading time":"4 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/physicscatalyst.com\/article\/integration-of-1-a2-x2\/#article","isPartOf":{"@id":"https:\/\/physicscatalyst.com\/article\/integration-of-1-a2-x2\/"},"author":{"name":"physicscatalyst","@id":"https:\/\/physicscatalyst.com\/article\/#\/schema\/person\/9b302efdc9b32e459cb1e61ab7506d3f"},"headline":"integration of 1\/a^2+ x^2, 1\/a^2- x^2, 1\/x^2 -a^2","datePublished":"2024-01-28T13:45:56+00:00","dateModified":"2024-01-28T13:46:35+00:00","mainEntityOfPage":{"@id":"https:\/\/physicscatalyst.com\/article\/integration-of-1-a2-x2\/"},"wordCount":752,"commentCount":0,"publisher":{"@id":"https:\/\/physicscatalyst.com\/article\/#organization"},"articleSection":["Maths"],"inLanguage":"en-US","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/physicscatalyst.com\/article\/integration-of-1-a2-x2\/#respond"]}]},{"@type":"WebPage","@id":"https:\/\/physicscatalyst.com\/article\/integration-of-1-a2-x2\/","url":"https:\/\/physicscatalyst.com\/article\/integration-of-1-a2-x2\/","name":"integration of 1\/a^2+ x^2, 1\/a^2- x^2, 1\/x^2 -a^2 - physicscatalyst's Blog","isPartOf":{"@id":"https:\/\/physicscatalyst.com\/article\/#website"},"datePublished":"2024-01-28T13:45:56+00:00","dateModified":"2024-01-28T13:46:35+00:00","breadcrumb":{"@id":"https:\/\/physicscatalyst.com\/article\/integration-of-1-a2-x2\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/physicscatalyst.com\/article\/integration-of-1-a2-x2\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/physicscatalyst.com\/article\/integration-of-1-a2-x2\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/physicscatalyst.com\/article\/"},{"@type":"ListItem","position":2,"name":"Maths","item":"https:\/\/physicscatalyst.com\/article\/maths\/"},{"@type":"ListItem","position":3,"name":"integration of 1\/a^2+ x^2, 1\/a^2- x^2, 1\/x^2 -a^2"}]},{"@type":"WebSite","@id":"https:\/\/physicscatalyst.com\/article\/#website","url":"https:\/\/physicscatalyst.com\/article\/","name":"physicscatalyst's Blog","description":"Learn free for class 9th, 10th science\/maths , 12th and IIT-JEE Physics and maths.","publisher":{"@id":"https:\/\/physicscatalyst.com\/article\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/physicscatalyst.com\/article\/?s={search_term_string}"},"query-input":{"@type":"PropertyValueSpecification","valueRequired":true,"valueName":"search_term_string"}}],"inLanguage":"en-US"},{"@type":"Organization","@id":"https:\/\/physicscatalyst.com\/article\/#organization","name":"physicscatalyst","url":"https:\/\/physicscatalyst.com\/article\/","logo":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/physicscatalyst.com\/article\/#\/schema\/logo\/image\/","url":"https:\/\/physicscatalyst.com\/article\/wp-content\/uploads\/2024\/08\/cropped-logo-1.jpg","contentUrl":"https:\/\/physicscatalyst.com\/article\/wp-content\/uploads\/2024\/08\/cropped-logo-1.jpg","width":96,"height":96,"caption":"physicscatalyst"},"image":{"@id":"https:\/\/physicscatalyst.com\/article\/#\/schema\/logo\/image\/"},"sameAs":["https:\/\/www.facebook.com\/PhysicsCatalyst","https:\/\/x.com\/physicscatalyst","https:\/\/www.youtube.com\/user\/thephysicscatalyst","https:\/\/www.instagram.com\/physicscatalyst\/"]},{"@type":"Person","@id":"https:\/\/physicscatalyst.com\/article\/#\/schema\/person\/9b302efdc9b32e459cb1e61ab7506d3f","name":"physicscatalyst","sameAs":["https:\/\/physicscatalyst.com","https:\/\/www.facebook.com\/PhysicsCatalyst","https:\/\/x.com\/physicscatalyst"]}]}},"uagb_featured_image_src":{"full":false,"thumbnail":false,"medium":false,"medium_large":false,"large":false,"1536x1536":false,"2048x2048":false,"shareaholic-thumbnail":false},"uagb_author_info":{"display_name":"physicscatalyst","author_link":"https:\/\/physicscatalyst.com\/article\/author\/physicscatalyst\/"},"uagb_comment_info":0,"uagb_excerpt":"Integration of 1\/a^2+ x^2, 1\/a^2- x^2, 1\/x^2 -a^2 can be obtained using trigonometric substitution, integration by partial fractions. The formula given are $\\int \\frac {1}{x^2 + a^2} dx = \\frac {1}{a} \\tan ^{-1} (\\frac {x}{a}) + C$ $\\int \\frac {1}{x^2 – a^2} dx = \\frac {1}{2a} ln |\\frac {x-a}{x+a}| + C$ $\\int \\frac {1}{a^2 –…","_links":{"self":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/8930","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/comments?post=8930"}],"version-history":[{"count":3,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/8930\/revisions"}],"predecessor-version":[{"id":8933,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/8930\/revisions\/8933"}],"wp:attachment":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/media?parent=8930"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/categories?post=8930"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/tags?post=8930"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}