Question Number 55273 by maxmathsup by imad last updated on 20/Feb/19
$${let}\:\:{f}\left({x}\right)\:=\int_{{x}^{\mathrm{2}} } ^{\mathrm{1}+{x}} \:\:\frac{{dt}}{\mathrm{1}+{t}+{t}^{\mathrm{2}} } \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({x}\right)\:{interms}\:{of}\:{x} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)\:{and}\:{lim}_{{x}\rightarrow+\infty} \:\:{f}\left({x}\right) \\ $$
Commented by maxmathsup by imad last updated on 21/Feb/19
![1) we have f(x)=∫_x^2 ^(1+x) (dt/(t^2 +2 (t/2)+(1/4)+(3/4))) =∫_x^2 ^(1+x) (dt/((t+(1/2))^(2 ) +(3/4))) =_(t+(1/2)=((√3)/2)u) (4/3)∫_((2x^2 +1)/(√3)) ^((2(1+x)+1)/(√3)) (1/(1+u^2 )) ((√3)/2) du =(2/(√3))[arctan(u)]_((2x^2 +1)/(√3)) ^((2x+3)/(√3)) ⇒ f(x)=(2/(√3)){ arctan(((2x+3)/(√3)))−arctan(((2x^2 +1)/(√3)))} 2)lim_(x→0) f(x)=(2/(√3)){ arctan((√3)) −arctan((1/(√3)))} =(2/(√3)){ (π/3) −(π/6)} =(2/(√3)){(π/6)} =(π/(3(√3))) lim_(x→+∞) f(x) =(2/(√3)){ arctan(+∞)−arctan(+∞)}=(2/(√3)){(π/2) −(π/2)}=0](Q55345.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)=\int_{{x}^{\mathrm{2}} } ^{\mathrm{1}+{x}} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{2}\:\frac{{t}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}}\:=\int_{{x}^{\mathrm{2}} } ^{\mathrm{1}+{x}} \:\:\frac{{dt}}{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}\:} \:+\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$=_{{t}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{u}} \:\:\:\:\:\frac{\mathrm{4}}{\mathrm{3}}\int_{\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{2}\left(\mathrm{1}+{x}\right)+\mathrm{1}}{\sqrt{\mathrm{3}}}} \:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{du}\:=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\left[{arctan}\left({u}\right)\right]_{\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{2}{x}+\mathrm{3}}{\sqrt{\mathrm{3}}}} \:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\left\{\:{arctan}\left(\frac{\mathrm{2}{x}+\mathrm{3}}{\sqrt{\mathrm{3}}}\right)−{arctan}\left(\frac{\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\right\} \\ $$$$\left.\mathrm{2}\right){lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\left\{\:{arctan}\left(\sqrt{\mathrm{3}}\right)\:−{arctan}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\right\} \\ $$$$=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\left\{\:\frac{\pi}{\mathrm{3}}\:−\frac{\pi}{\mathrm{6}}\right\}\:=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\left\{\frac{\pi}{\mathrm{6}}\right\}\:=\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)\:=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\left\{\:{arctan}\left(+\infty\right)−{arctan}\left(+\infty\right)\right\}=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\left\{\frac{\pi}{\mathrm{2}}\:−\frac{\pi}{\mathrm{2}}\right\}=\mathrm{0} \\ $$