Question-197226

Question Number 197226 by sonukgindia last updated on 10/Sep/23

Answered by Frix last updated on 10/Sep/23

∫(dx/(1+(√(x^2 +x+1)))) =^(t=((2x+1+2(√(x^2 +x+1)))/( (√3)))) =∫((t^2 +1)/(t(t^2 +((4t)/( (√3)))+1)))dt=∫((1/t)+(2/(t+(√3)))−(6/(3t+(√3))))dt= =ln t +2ln (t+(√3)) −2ln (3t+(√3)) = =ln ((t(t+(√3))^2 )/((3t+(√3))^2 )) = =ln ((x−1)(x+2)(2x+1)+2(√((x^2 +x+1)^3 ))) −ln x^2 +C

$$\int\frac{{dx}}{\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}\:\overset{{t}=\frac{\mathrm{2}{x}+\mathrm{1}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{\:\sqrt{\mathrm{3}}}} {=} \\ $$$$=\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}\left({t}^{\mathrm{2}} +\frac{\mathrm{4}{t}}{\:\sqrt{\mathrm{3}}}+\mathrm{1}\right)}{dt}=\int\left(\frac{\mathrm{1}}{{t}}+\frac{\mathrm{2}}{{t}+\sqrt{\mathrm{3}}}−\frac{\mathrm{6}}{\mathrm{3}{t}+\sqrt{\mathrm{3}}}\right){dt}= \\ $$$$=\mathrm{ln}\:{t}\:+\mathrm{2ln}\:\left({t}+\sqrt{\mathrm{3}}\right)\:−\mathrm{2ln}\:\left(\mathrm{3}{t}+\sqrt{\mathrm{3}}\right)\:= \\ $$$$=\mathrm{ln}\:\frac{{t}\left({t}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }{\left(\mathrm{3}{t}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\:= \\ $$$$=\mathrm{ln}\:\left(\left({x}−\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)+\mathrm{2}\sqrt{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{3}} }\right)\:−\mathrm{ln}\:{x}^{\mathrm{2}} \:+{C} \\ $$

Facebook Tweet Pin

Question-197226

Leave a Reply Cancel reply