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• 06:19, 29 November 2022 Jorgen70034@students.laalliance.org talk contribs created page 7.1 Integration By Parts/26 (Created page with "<math>\int_{1}^{\sqrt{3}}arctan\left(\frac{1}{x}\right)dx</math> <math>letu =arctan\left(\frac{1}{x}\right)du=dx=du=\frac{1}{1+(1/x)^2}x\frac{-1}{x^2}dx=\frac{-dx}{x^2+1}</math> <math>\int_{1}^{\sqrt{3}}arctan\frac{1}{x}dx=[xarctan\frac{1}{x}]\bigg|_{0}^{1}+\int_{1}^{\sqrt{3}}\frac{x}{dx}x^2+1=\sqrt{3}\frac{\pi}{6}=\frac{1}{4}=\frac{1}{2}[in(x^2+1)]\bigg|_{1}^{\sqrt{3}}</math> <math>=\frac{\pi\sqrt{3}}{6}-\frac{\pi}{4}(in4-in2)=\frac{\pi\sqrt{3}}{6}-\frac{\pi}{2}=\fr...")
• 00:24, 29 November 2022 Jorgen70034@students.laalliance.org talk contribs created page 6.2 Volumes/29 (Created page with "<math>y^2+R=1</math> <math>\int_{0}^{1}\pi(1-y^2)^2-\pi(1-\sqrt[31]{y})^2dy=\pi\sqrt{\int_{0}^{1}}1-2y^2+y^4dy</math> <math>\int_{0}^{1}1-2-\sqrt[3]{y}</math>")
• 19:42, 1 September 2022 Jorgen70034@students.laalliance.org talk contribs created page 5.5 The Substitution Rule/63 (Created page with "<math>\int_{a}^{0}\{x}sqrt{x^2+a^2}dx</math>")
• 16:00, 23 August 2022 Jorgen70034@students.laalliance.org talk contribs created page Jorge (Created page with "hello")
• 15:50, 23 August 2022 User account Jorgen70034@students.laalliance.org talk contribs was created automatically