7.1 Integration By Parts/26: Revision history

29 November 2022

• curprev 06:1906:19, 29 November 2022‎ Jorgen70034@students.laalliance.org talk contribs‎ 554 bytes +554‎ 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..."