We are tasked with evaluating the definite integral:\u222b01ex1+e2x\u2009dx\\int_0^1 \\frac{e^x}{1 + e^{2x}} \\, dx\u222b01\u200b1+e2xex\u200bdx<\/p>\n\n\n\n
To solve this, we first observe the structure of the integrand. Let’s proceed step-by-step:<\/p>\n\n\n\n
Step 1: Simplifying the integrand<\/h3>\n\n\n\n
The given expression is:ex1+e2x\\frac{e^x}{1 + e^{2x}}1+e2xex\u200b<\/p>\n\n\n\n
We can make a substitution to simplify the expression. Let:u=exu = e^xu=ex<\/p>\n\n\n\n
Then, the differential of uuu is:du=ex\u2009dxdu = e^x \\, dxdu=exdx<\/p>\n\n\n\n
When x=0x = 0x=0, u=e0=1u = e^0 = 1u=e0=1, and when x=1x = 1x=1, u=e1=eu = e^1 = eu=e1=e.<\/p>\n\n\n\n
Thus, the limits of integration change from x=0x = 0x=0 to x=1x = 1x=1, becoming u=1u = 1u=1 to u=eu = eu=e.<\/p>\n\n\n\n
Step 2: Substitute into the integral<\/h3>\n\n\n\n
Now, substitute u=exu = e^xu=ex and du=ex\u2009dxdu = e^x \\, dxdu=exdx into the integral:\u222b01ex1+e2x\u2009dx=\u222b1e11+u2\u2009du\\int_0^1 \\frac{e^x}{1 + e^{2x}} \\, dx = \\int_1^e \\frac{1}{1 + u^2} \\, du\u222b01\u200b1+e2xex\u200bdx=\u222b1e\u200b1+u21\u200bdu<\/p>\n\n\n\n
Step 3: Recognizing the standard integral<\/h3>\n\n\n\n
The integral:\u222b11+u2\u2009du\\int \\frac{1}{1 + u^2} \\, du\u222b1+u21\u200bdu<\/p>\n\n\n\n
is a standard form, and its solution is:arctan\u2061(u)\\arctan(u)arctan(u)<\/p>\n\n\n\n
Thus, our integral becomes:\u222b1e11+u2\u2009du=arctan\u2061(u)\u22231e\\int_1^e \\frac{1}{1 + u^2} \\, du = \\arctan(u) \\bigg|_1^e\u222b1e\u200b1+u21\u200bdu=arctan(u)\u200b1e\u200b<\/p>\n\n\n\n
Step 4: Evaluate the definite integral<\/h3>\n\n\n\n
Now, evaluate the arctangent at the bounds u=eu = eu=e and u=1u = 1u=1:arctan\u2061(e)\u2212arctan\u2061(1)\\arctan(e) – \\arctan(1)arctan(e)\u2212arctan(1)<\/p>\n\n\n\n
We know that:arctan\u2061(1)=\u03c04\\arctan(1) = \\frac{\\pi}{4}arctan(1)=4\u03c0\u200b<\/p>\n\n\n\n
Thus, the integral becomes:arctan\u2061(e)\u2212\u03c04\\arctan(e) – \\frac{\\pi}{4}arctan(e)\u22124\u03c0\u200b<\/p>\n\n\n\n
Final Answer<\/h3>\n\n\n\n
The value of the definite integral is:arctan\u2061(e)\u2212\u03c04\\arctan(e) – \\frac{\\pi}{4}arctan(e)\u22124\u03c0\u200b<\/p>\n\n\n\n
This is the exact answer for the given definite integral.<\/p>\n\n\n\n Evaluate the definite integral. 1 ex 1 + e2x dx 0 The Correct Answer and Explanation is: We are tasked with evaluating the definite integral:\u222b01ex1+e2x\u2009dx\\int_0^1 \\frac{e^x}{1 + e^{2x}} \\, dx\u222b01\u200b1+e2xex\u200bdx To solve this, we first observe the structure of the integrand. Let’s proceed step-by-step: Step 1: Simplifying the integrand The given expression is:ex1+e2x\\frac{e^x}{1 + e^{2x}}1+e2xex\u200b […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-42318","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42318","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=42318"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42318\/revisions"}],"predecessor-version":[{"id":42320,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42318\/revisions\/42320"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=42318"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=42318"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=42318"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}![\"\"](\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-297.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"