{"id":97870,"date":"2021-10-20T10:35:38","date_gmt":"2021-10-20T15:35:38","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=97870"},"modified":"2026-03-28T11:46:35","modified_gmt":"2026-03-28T16:46:35","slug":"integration-by-parts","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/integration-by-parts\/","title":{"rendered":"Integration by Parts"},"content":{"rendered":"<h1>Integration by Parts<\/h1>\n\n\t\t\t<div id=\"mmDeferVideoEncompass_L5CabDQ7JCE\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_L5CabDQ7JCE\" data-source-videoID=\"L5CabDQ7JCE\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Integration by Parts Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Integration by Parts\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_L5CabDQ7JCE:hover {cursor:pointer;} img#videoThumbnailImage_L5CabDQ7JCE {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/1767-thumb-final-2.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_L5CabDQ7JCE\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_L5CabDQ7JCE\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Integration by Parts\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_L5CabDQ7JCE\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_L5CabDQ7JCE\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_L5CabDQ7JCE\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hey guys! Welcome to this video on <strong>integration<\/strong> by parts! At this point, you should know how to integrate basic functions, including some polynomials and other problems that can be solved with \\(u\\)-substitution. However, not all functions fall into these categories, and today we will discuss another method of integration that will enable us to solve more integration problems. This method is called integration by parts.<\/p>\n<p>To begin, let\u2019s take a look at a problem we should already be able to solve.<\/p>\n<div class=\"examplesentence\">\\(\\int \\left [ x+\\sin \\left ( 2x \\right )\\right ]dx\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>First, we can break this into two separate integrals by using the addition rule. So the addition rule says that we can break it up where there&#8217;s a plus sign, so we&#8217;ll have:<\/p>\n<div class=\"examplesentence\">\\(\\int x~dx+\\int \\sin\\left ( 2x \\right )dx\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe first integral, \\(\\int x~dx\\), is simple to solve. Just increase the power by 1, and divide by the new exponent. So this part is equal to, if you increase \\(x^{1}\\) by 1, you\u2019ll get \\(x^{2}\\), and then divide by 2, is the same as multiplying by \\(\\frac{1}{2}\\), plus \\(c\\), our integration constant.<\/p>\n<div class=\"examplesentence\">\\(\\int x~ dx=\\frac{1}{2}x^{2}+c\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>For the second integral, notice that the derivative of \\(2x\\) is 2, which is a scalar multiple of the implied 1 in front of the sine function. Because of this relationship, we can apply \\(u\\)-substitution to solve the second integral.<\/p>\n<div class=\"examplesentence\">\\(\\int \\sin\\left ( 2x \\right )dx\\rightarrow \\int \\frac{1}{2}\\sin\\left ( u \\right )du\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>So, if we have \\(\\int \\sin(2x)dx\\), and we call \\(2x\\) our \\(u\\), then \\(du=2~dx\\). So this will be equal to \\(\\int \\sin(u)du\\).<\/p>\n<div class=\"examplesentence\">\\(u=2x\\)<br \/>\n\\(du=2~dx\\)<\/div>\n<p>\n&nbsp;<\/p>\n<div class=\"examplesentence\">\\(\\int \\sin\\left ( 2x \\right )dx=\\int \\sin\\left ( u \\right )du\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>But we\u2019re actually going to have to use \\(\\frac{1}{2}du\\)  to get it to be equal to \\(dx\\). We just divided by 2 on both sides. <\/p>\n<div class=\"examplesentence\">\\(\\frac{1}{2}du=dx\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>And since we have a \\(dx\\) here, that\u2019s what we\u2019re going to want to do. So we\u2019re going to write \\(\\frac{1}{2}\\) right here:<\/p>\n<div class=\"examplesentence\">\\(\\int \\sin\\left ( 2x \\right )dx=\\int \\frac{1}{2}\\sin\\left ( u \\right )du\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Now we integrate this, and we\u2019ll get:<\/p>\n<div class=\"examplesentence\">\\(-\\frac{1}{2}\\cos\\left ( u \\right )+c\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Because \\(-\\cos(u)\\) is the integral of \\(\\sin(u)\\). Now all we have to do is plug in our \\(u\\), which is \\(2x\\), and we get:<\/p>\n<div class=\"examplesentence\">\\(-\\frac{1}{2}\\cos\\left ( 2x \\right )+c\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Now, all we have to do is put our addition terms back together. So this is equal to our first part, \\(\\frac{1}{2}x^{2}\\). We\u2019re going to leave the \\(c\\) alone for a second. And then \\(-\\frac{1}{2}\\cos(2x)+c\\).<\/p>\n<div class=\"examplesentence\">\\(\\int x~ dx+\\int \\sin\\left ( 2x \\right )dx\\)\\(=\\frac{1}{2}x^{2}-\\frac{1}{2}\\cos\\left ( 2x \\right )+c\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Don\u2019t be concerned by the fact that one of our \\(c\\)\u2019s went away. When we add together our two integration constants from the first and second terms, they combine to form another constant, so we are safe to lump them together using just one letter.<\/p>\n<p>Solving that problem wasn\u2019t so bad, but consider this next problem:<\/p>\n<div class=\"examplesentence\">\\(\\int x\\sin\\left ( 2x \\right )dx\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>\\(\\int x\\sin(2x)~dx\\) is almost identical to the one we just solved, except now we are dealing with multiplication where before we were simply adding the two terms. Unfortunately, we can\u2019t just break this integral into two separate integrals like we can with the addition rule. What can we do, then?<\/p>\n<p>Think back to what we know about taking derivatives of products. We had a special rule, the product rule, to help us with those types of problems. Similarly, with integration, we have a defined process for handling products, and it was actually derived by mathematicians from the product rule for derivatives. This process, as you might have guessed, is called integration by parts, and it goes like this.<\/p>\n<p>If you have an integral that can be written in the form \\(\\int u~dv\\), then its solution is equal to \\(uv-\\int v~ du\\).<\/p>\n<div class=\"examplesentence\">\\(\\int u~ dv=uv-\\int v~ du\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>At first glance, this may seem a bit confusing or unhelpful, but let\u2019s break it down and then apply it to our example.<\/p>\n<p>To begin, there are two functions, \\(u\\) and \u2009\\(dv\\), whose product we are interested in taking the integral of. If we know how to take the integral of \u2009\\(dv\\), and we know how to take the derivative of \\(u\\), then we can use a little bit of algebra to apply the integration by parts formula and determine our answer.<\/p>\n<p>Consider again our example: \\(\\int x\\sin\\left ( 2x \\right )dx\\)<\/p>\n<p>The two functions here are \\(x\\) and \\(\\sin2x~dx\\), so we will assign them to be our \\(u\\) and \u2009\\(dv\\), respectively.<\/p>\n<div class=\"examplesentence\">\\(u=x\\)<br \/>\n\\(\u2009dv=\\sin\\left ( 2x \\right )dx\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Now, notice that I kept the \\(dx\\) with the \\(dv\\). This is important because we\u2019re going to integrate this part later on.<\/p>\n<p>In order to determine our solution using the integration by parts formula, we need to calculate \\(uv\\) and \\(\\int v~du\\). We know that \\(u=x\\), but what is \\(v\\)? To find \\(v\\), we just need to take the antiderivative of \\(dv\\).<\/p>\n<div class=\"examplesentence\">\\(v=\\int \\sin\\left ( 2x \\right )dx\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, remember, we found this earlier, so this was equal to \\(-\\frac{1}{2}\\cos(2x)\\).<\/p>\n<div class=\"examplesentence\">\\(v=\\int \\sin(2x)dx=-\\frac{1}{2}\\cos(2x)\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>And we\u2019re not going to write the \\(+c\\) yet because we\u2019ll do that at the end. Since we\u2019re integrating a whole indefinite integral, we\u2019ll do that at the very end. We don\u2019t have to worry about it in our little steps in between.<\/p>\n<p>So now we can calculate \\(uv\\).<\/p>\n<div class=\"examplesentence\">\\(uv=x\\left [ -\\frac{1}{2}\\cos(2x) \\right ]\\)\\(=-\\frac{1}{2}x\\cos(2x)\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>To find the last piece, \\(\\int v~ du\\), we need to compute \\(du\\) and then integrate. The derivative of \\(u\\) is the derivative of \\(x\\), which is 1, so \\(du\\) is equal to\u2009\u2009\\(dx\\).<\/p>\n<div class=\"examplesentence\">\\(du=\u2009\u2009dx\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Putting this together:<\/p>\n<div class=\"examplesentence\">\\(\\int v~ du=\\int -\\frac{1}{2}\\cos(2x)dx\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Let\u2019s put together what we have so far. Remember, the formula tells us that<\/p>\n<div class=\"examplesentence\">\\(\\int u~ dv=uv-\\int v~ du\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>We assigned \\(u=x\\), and \\(dv=\\sin(2x)\u2009dx\\), which made \\(du=dx\\), and \\(v=-\\frac{1}{2}\\cos(2x)\\).<\/p>\n<div class=\"examplesentence\">\\(u = x\\)\t\t\t\\(du = \u2009dx\\)<br \/>\n\\(v=-\\frac{1}{2}\\cos(2x)\\)\t\\(dv=\\sin(2x)\u2009dx\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Putting these pieces together into the formula, we now have that<\/p>\n<div class=\"examplesentence\">\\(uv-\\int v~ du\\)\\(=-\\frac{1}{2}x\\cos(2x)\\)\\(-\\int \\left [ -\\frac{1}{2}\\cos(2x) \\right ]dx\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>All we need to do from here is compute the second integral and simplify the entire expression. Let\u2019s move that \\(-\\frac{1}{2}\\) outside of the integral, which will cancel out the negative sign.<\/p>\n<p>So this first part\u2019s going to stay the same And we can pull this \\(-\\frac{1}{2}\\) out, so we\u2019ll have minus a negative, which becomes plus.<\/p>\n<div class=\"examplesentence\">\\(-\\frac{1}{2}x\\cos(2x)+\\frac{1}{2}\\int \\cos(2x)dx\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Now, the antiderivative of cosine is sine, and we pull out a \\(\\frac{1}{2}\\) because of this \\(2x\\) in here. So that will look like this. Again, this first part stays the same, and \\(\\int \\cos(2x)dx=\\frac{1}{2}\\sin{2x}\\).<\/p>\n<div class=\"examplesentence\"><span style=\"font-size: 90%\">\\(-\\frac{1}{2}x\\cos(2x)+\\frac{1}{2}(\\frac{1}{2})\\sin(2x)+c\\)<\/span><\/div>\n<p>\n&nbsp;<\/p>\n<p>And this is where we\u2019re going to add our \\(+c\\), because we\u2019re at the very end and we just need to simplify.<\/p>\n<p>So now, all we have to do is simplify, and we do that by, again, keeping this first part the same. There\u2019s nothing we can simplify in this part. And then, multiplying \\(\\frac{1}{2}\\times \\frac{1}{2}=\\frac{1}{4}\\).<\/p>\n<div class=\"examplesentence\">\\(-\\frac{1}{2}x\\cos(2x)+\\frac{1}{4}\\sin\\left ( 2x \\right )+c\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>So our final answer is:<\/p>\n<div class=\"examplesentence\"><span style=\"font-size: 90%\">\\(\\int x\\sin(2x)dx\\)\\(=-\\frac{1}{2}x\\cos(2x)+\\frac{1}{4}\\sin(2x)+c\\)<\/span><\/div>\n<p>\n&nbsp;<\/p>\n<p>This process may seem long, but with practice it will come along without much difficulty.<\/p>\n<p>At the beginning of this problem, we decided to let \\(u=x\\), and to let \\(dv=\\sin(2x) dx\\). This enabled us to solve the integral using the integration by parts formula. However, if we would have tried to solve the integral with these switched (that is, \\(u=\\sin(2x)\\), and \\(dv=x~ dx\\)), we would have run into some problems. The new integral, \\(\\int v~ du\\), would have been \\(\\int x^{2}\\cos(2x)dx\\). Like the problem we started with, we cannot easily solve this new integral with previous methods. By choosing incorrectly, we effectively go in circles.<\/p>\n<p>Clearly, it is important to choose correctly which functions will be \\(u\\) and which will be \\(dv\\) when we integrate by parts. But how can we know the correct choice, apart from trial and error? Some mathematicians have developed a method for making this choice, called \u201cILATE.\u201d ILATE is an acronym to help us choose our u, based on an order of priority.<\/p>\n<p>When approaching a problem for which you will need to use integration by parts, first check to see if there is any\u2026<\/p>\n<div class=\"examplesentence\">I, inverse trig function. If so, select this to be \\(u\\). If not, then look for an\u2026<br \/>\n&nbsp;<br \/>\nL, logarithmic function. If there is one, then it will be what you choose for \\(u\\). Otherwise, check for an\u2026<br \/>\n&nbsp;<br \/>\nA, algebraic expression, such as \\(x\\) or \\(x^2\\). If there is none, see if there is any\u2026<br \/>\n&nbsp;<br \/>\nT, trigonometric functions, like \\(\\sin(x)\\) or \\(\\cos(x)\\). If none of the previous expressions are in the problem, then finally, check for\u2026<br \/>\n&nbsp;<br \/>\nE, exponential functions, such as \\(e^x\\).<\/div>\n<p>\nWhichever of these you find in order of ILATE, that will be your \\(u\\). The part of the integral that is not chosen to be \\(u\\) will be your \\(dv\\).<\/p>\n<p>Let\u2019s look back to our problem from earlier one more time to apply ILATE. <\/p>\n<div class=\"examplesentence\">\\(\\int x\\sin(2x)dx\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>First, are there any (I) inverse trig functions? No, there are not. Next is L. Are there any log functions present? Again, no. Are there any (A) algebraic expressions? Yes! By using ILATE, we see that \\(x\\) is the best choice for our \\(u\\), because it is an algebraic expression. Notice that the sine function is a (T) trig function, but we don\u2019t choose that to be u because the T comes after A in ILATE.<\/p>\n<p>Let\u2019s work one more problem.<\/p>\n<div class=\"examplesentence\">\\(\\int x^{2}\\ln(x)dx\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>We can\u2019t use simple antiderivatives or \\(u\\)-substitution to solve this integral, and because we have a product of two functions, \\(x^{2}\\) and \\(\\ln(x)\\), we\u2019re going to use integration by parts. Which of the two will be our \\(u\\), and which will be our \\(dv\\)? We will use ILATE to decide.<\/p>\n<p>There are no (I) inverse trig functions. There is, however, an (L) log function. We assign \\(u=\\ln(x)\\), and we have no need to continue with A, T, and E. The leftover portion of this integral, \\(x^{2}~dx\\), is our \\(dv\\).<\/p>\n<div class=\"examplesentence\">\\(u=\\ln(x)\\)<br \/>\n\\(dv=x^{2}\u2009dx\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>To use the integration by parts formula, let\u2019s go ahead and determine what \\(du\\) and \\(v\\) are, based on our choice. So \\(du\\) is the derivative of \\(\\ln(x)\\), which is equal to \\(\\frac{1}{x}\u2009dx\\), and \\(v=\\int x^{2}~dx\\), which is equal to \\(\\frac{1}{3}x^{3}\\). Remember, we don\u2019t have to put the \\(+c\\) yet. We\u2019ll put that at the very end. <\/p>\n<div class=\"examplesentence\">\\(du=\\frac{1}{x}dx\\)<br \/>\n\\(v=\\frac{1}{3}x^{3}\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Because the formula for integration by parts is:<\/p>\n<div class=\"examplesentence\">\\(\\int u~ dv=uv-\\int v~ du\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>We plug in our substitutions and get this. So \\(uv=\\ln(x)\\frac{1}{3}x^{3}\\), so I\u2019m going to write the \\(\\frac{1}{3}x^{3}\\) in front (that\u2019s just the more formal way to write it), then \\(-\\int v~ du\\).<\/p>\n<div class=\"examplesentence\">\\(\\frac{1}{3}x^{3}ln(x)-\\int v~ du\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>So, \\(\\int \\frac{1}{3}x^{3}\\times du\\), which is \\(du=\\frac{1}{x} dx\\).<\/p>\n<div class=\"examplesentence\">\\(\\frac{1}{3}x^{3}\\ln(x)-\\int \\frac{1}{3}x^{3}(\\frac{1}{x})dx\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Now we\u2019re going to go ahead and simplify. So this is equal to:<\/p>\n<div class=\"examplesentence\">\\(\\frac{1}{3}x^{3}ln(x)-\\int \\frac{1}{3}x^{2}~dx\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>The next step is that we\u2019ll pull out our \\(\\frac{1}{3}\\). So again, this first part stays the same, and we can subtract \\(\\frac{1}{3}\\int x^{2}~dx\\).<\/p>\n<div class=\"examplesentence\">\\(\\frac{1}{3}x^{3}\\ln(x)-\\frac{1}{3}\\int x^{2}~dx\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>And then we can integrate \\(x^{2}\\) to get \\(\\frac{1}{3}x^{3}\\). And this is where we\u2019ll add our \\(+c\\).<\/p>\n<div class=\"examplesentence\">\\(\\frac{1}{3}x^{3}\\ln(x)-\\frac{1}{3}\\left ( \\frac{1}{3x}x^{3} \\right )+c\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>And now all we have to do is simplify this last part right here. So we have<\/p>\n<div class=\"examplesentence\">\\(\\frac{1}{3}x^{3}\\ln(x)-\\frac{1}{9}x^{3}+c\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>And that\u2019s our final answer!<\/p>\n<p>Let\u2019s briefly run back through everything we\u2019ve talked about. First, not all integrals can be solved using basic antiderivatives and u-substitution. Integration by parts allows us to work out more integrals, specifically those containing products of functions, using the formula<\/p>\n<div class=\"examplesentence\">\\(\\int u~ dv=uv-\\int v~ du\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>To make the best choice for which function should be \\(u\\), we use the acronym ILATE: inverse trig functions, log functions, algebraic expressions, trig functions, and finally, exponential functions. The part of the original integrand that we do not select to be \\(u\\) will be \\(dv\\). Before plugging everything into the formula, we need to calculate \\(du\\), the derivative of \\(u\\), and \\(v\\), the antiderivative of \\(dv\\). Once we have all four of those pieces, we simply substitute them all into the formula and simplify.<\/p>\n<p>With a little practice, you\u2019ll be fluent with integration by parts in no time!<\/p>\n<p>Thanks for watching, and happy studying!<\/p>\n<\/div>\n<\/div>\n\n<p>&nbsp;<\/p>\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/calculus\/\"><strong>Return to Calculus Videos<\/strong><\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Integration by Parts &nbsp; Return to Calculus Videos<\/p>\n","protected":false},"author":22,"featured_media":97873,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-97870","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-calculus-videos","7":"page_category-video-pages-for-study-course-sidebar-ad","8":"page_type-video","9":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/97870","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/22"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=97870"}],"version-history":[{"count":7,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/97870\/revisions"}],"predecessor-version":[{"id":281069,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/97870\/revisions\/281069"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/97873"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=97870"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}