Question Number 226464 by Spillover last updated on 29/Nov/25
Commented by Frix last updated on 29/Nov/25
$$\mathrm{By}\:\mathrm{parts} \\ $$$${u}'=\mathrm{1}\:\rightarrow\:{u}={x} \\ $$$${v}=\mathrm{cot}^{−\mathrm{1}} \:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:\rightarrow\:{v}'=−\frac{\mathrm{2}{x}−\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)} \\ $$$$\Rightarrow \\ $$$$\int\mathrm{cot}^{−\mathrm{1}} \:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:{dx}= \\ $$$$={x}\mathrm{cot}^{−\mathrm{1}} \:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:+\int\frac{{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}{dx} \\ $$$$…\mathrm{which}\:\mathrm{should}\:\mathrm{be}\:\mathrm{easy}… \\ $$
Commented by Spillover last updated on 30/Nov/25
$${appriciate} \\ $$