Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 172823 by Mikenice last updated on 01/Jul/22

Answered by thfchristopher last updated on 03/Jul/22

$${\int }_0^1{\cot }^{-1}(x^2-x+1)dx$$$$={\left[x{\cot }^{-1}(x^2-x+1)\right]}_0^1+{\int }_0^1\frac{x(2x-1)}{{(x^2-x+1)}^2+1}dx$$$$=\frac{\pi }{4}+{\int }_0^1\frac{2x^2-x}{x^4-2x^3+3x^2-2x+2}dx$$$$=\frac{\pi }{4}+{\int }_0^1\frac{2x^2-x}{(x^2-2x+2)(x^2+1)}dx$$$$\frac{2x^2-x}{(x^2-2x+2)(x^2+1)}\equiv \frac{Ax+B}{x^2-2x+2}+\frac{Cx+D}{x^2+1}$$$$\mathrm{By}\mathrm{solving},$$$$A=1,B=0,C=-1,D=0$$$$\therefore \frac{2x^2-x}{(x^2-2x+2)(x^2+1)}=\frac{x}{x^2-2x+2}-\frac{x}{x^2+1}$$$${\int }_0^1\frac{2x^2-x}{(x^2-2x+2)(x^2+1)}dx$$$$={\int }_0^1\frac{x}{x^2-2x+2}dx-{\int }_0^1\frac{x}{x^2+1}dx$$$${\int }_0^1\frac{x}{x^2-2x+2}dx$$$$=\frac{1}{2}{\int }_0^1\frac{2x-2}{x^2-2x+2}dx+{\int }_0^1\frac{dx}{x^2-2x+2}$$$$={\left[\frac{1}{2}\ln (x^2-2x+2)\right]}_0^1+{\int }_0^1\frac{dx}{{(x-1)}^2+1}$$$$=-\frac{1}{2}\ln 2+{\left[{\tan }^{-1}(x-1)\right]}_0^1$$$$=\frac{\pi }{4}-\frac{1}{2}\ln 2$$$${\int }_0^1\frac{x}{x^2+1}dx$$$$=\frac{1}{2}{\int }_0^1\frac{2x}{x^2+1}dx$$$$={\left[\frac{1}{2}\ln (x^2+1)\right]}_0^1$$$$=\frac{1}{2}\ln 2$$$$\mathrm{Hence},{\int }_0^1{\cot }^{-1}(x^2-x+1)dx$$$$=\frac{\pi }{4}+\frac{\pi }{4}-\frac{1}{2}\ln 2-\frac{1}{2}\ln 2$$$$=\frac{\pi }{2}-\ln 2$$

Commented by Tawa11 last updated on 04/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com