Menu Close

eplcit-f-x-0-1-ln-x-t-t-2-dt-with-x-gt-1-4-2-calculate-0-1-ln-t-2-t-2-dt-

Tinku Tara 0 Comments
Pin




Question Number 73335 by mathmax by abdo last updated on 10/Nov/19
eplcit f(x)=∫_0 ^1 ln(x+t+t^2 )dt with x>(1/4) 2)calculate ∫_0 ^1 ln(t^2 +t +(√2))dt
$${eplcit}\:\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}+{t}+{t}^{\mathrm{2}} \right){dt}\:\:\:\:\:\:{with}\:{x}>\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({t}^{\mathrm{2}} \:+{t}\:+\sqrt{\mathrm{2}}\right){dt} \\ $$
Commented by mathmax by abdo last updated on 10/Nov/19
1) we have f(x)=∫_0 ^1 ln(t^2 +t+x)dt by parts we get f(x)=[tln(t^2 +t+x)]_0 ^1 −∫_0 ^1 t×((2t+1)/(t^2 +t +x))dt =ln(2+x)−∫_0 ^1 ((2t^2 +t)/(t^2 +t+x))dt we have ∫_0 ^1 ((2t^2 +t)/(t^2 +t +x))dt =∫_0 ^1 ((2(t^2 +t+x)−2t−2x+t)/(t^2 +t+x))dt =∫_0 ^1 (2−((t+2x)/(t^2 +t+x)))dt =2 −∫_0 ^1 ((t+2x)/(t^2 +t+x))dt =2−(1/2)∫_0 ^1 ((2t +4x+1−1)/(t^(2 ) +t+x))dt =2−(1/2)∫_0 ^1 ((2t+1)/(t^2 +t+x))dt−(1/2)∫_0 ^1 ((4x−1)/(t^2 +t+x))dt =2−(1/2)[ln(t^2 +t+x)]_0 ^1 −((4x−1)/2)∫_0 ^1 (dt/(t^2 +t+x)) =2−(1/2)(ln(2+x)−ln(x))−((4x−1)/2)∫_0 ^1 (dt/(t^2 +t+x)) ∫_0 ^1 (dt/(t^2 +t+x)) =∫_0 ^1 (dt/(t^2 +2(t/2)+(1/4)+x−(1/4))) =∫_0 ^1 (dt/((t+(1/2))^2 +((4x−1)/4))) (4x−1>0) =_(t+(1/2)=((√(4x−1))/2)u) (4/(4x−1)) ∫_(1/( (√(4x−1)))) ^(3/( (√(4x−1)))) (1/(1+u^2 )) ((√(4x−1))/2) du =(2/( (√(4x−1)))){ arctan((3/( (√(4x−1)))))−arctan((1/( (√(4x−1)))))} ⇒ f(x)=ln(2+x)−2+(1/2)ln(((2+x)/x))+((4x−1)/2)×(2/( (√(4x−1)))){arctan((3/( (√(4x−1)))) −arctan((1/( (√(4x−1)))))} f(x)=(3/2)ln(2+x)−(1/2)ln(x)−2 +(√(4x−1)){arctan((3/( (√(4x−1)))))−arctan((1/( (√(4x−1)))))}
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({t}^{\mathrm{2}} \:+{t}+{x}\right){dt}\:\:{by}\:{parts}\:{we}\:{get} \\ $$$${f}\left({x}\right)=\left[{tln}\left({t}^{\mathrm{2}} \:+{t}+{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} {t}×\frac{\mathrm{2}{t}+\mathrm{1}}{{t}^{\mathrm{2}} \:+{t}\:+{x}}{dt} \\ $$$$={ln}\left(\mathrm{2}+{x}\right)−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{t}^{\mathrm{2}} \:+{t}}{{t}^{\mathrm{2}} \:+{t}+{x}}{dt}\:\:{we}\:{have} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}{t}^{\mathrm{2}} \:+{t}}{{t}^{\mathrm{2}} \:+{t}\:+{x}}{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}\left({t}^{\mathrm{2}} \:+{t}+{x}\right)−\mathrm{2}{t}−\mathrm{2}{x}+{t}}{{t}^{\mathrm{2}} \:+{t}+{x}}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{2}−\frac{{t}+\mathrm{2}{x}}{{t}^{\mathrm{2}} \:+{t}+{x}}\right){dt}\:=\mathrm{2}\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}+\mathrm{2}{x}}{{t}^{\mathrm{2}} \:+{t}+{x}}{dt} \\ $$$$=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}{t}\:+\mathrm{4}{x}+\mathrm{1}−\mathrm{1}}{{t}^{\mathrm{2}\:} +{t}+{x}}{dt}\:=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}{t}+\mathrm{1}}{{t}^{\mathrm{2}} \:+{t}+{x}}{dt}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{4}{x}−\mathrm{1}}{{t}^{\mathrm{2}} +{t}+{x}}{dt} \\ $$$$=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left({t}^{\mathrm{2}} \:+{t}+{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\frac{\mathrm{4}{x}−\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{{t}^{\mathrm{2}} \:+{t}+{x}} \\ $$$$=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\mathrm{2}+{x}\right)−{ln}\left({x}\right)\right)−\frac{\mathrm{4}{x}−\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{{t}^{\mathrm{2}} \:+{t}+{x}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{{t}^{\mathrm{2}} +{t}+{x}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{2}\frac{{t}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+{x}−\frac{\mathrm{1}}{\mathrm{4}}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{4}{x}−\mathrm{1}}{\mathrm{4}}}\:\:\left(\mathrm{4}{x}−\mathrm{1}>\mathrm{0}\right) \\ $$$$=_{{t}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{4}{x}−\mathrm{1}}}{\mathrm{2}}{u}} \:\:\:\:\:\frac{\mathrm{4}}{\mathrm{4}{x}−\mathrm{1}}\:\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}{x}−\mathrm{1}}}} ^{\frac{\mathrm{3}}{\:\sqrt{\mathrm{4}{x}−\mathrm{1}}}} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{\sqrt{\mathrm{4}{x}−\mathrm{1}}}{\mathrm{2}}\:{du} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}{x}−\mathrm{1}}}\left\{\:{arctan}\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{4}{x}−\mathrm{1}}}\right)−{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}{x}−\mathrm{1}}}\right)\right\}\:\Rightarrow \\ $$$${f}\left({x}\right)={ln}\left(\mathrm{2}+{x}\right)−\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{2}+{x}}{{x}}\right)+\frac{\mathrm{4}{x}−\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}{x}−\mathrm{1}}}\left\{{arctan}\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{4}{x}−\mathrm{1}}}\right.\right. \\ $$$$\left.−{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}{x}−\mathrm{1}}}\right)\right\} \\ $$$${f}\left({x}\right)=\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\mathrm{2}+{x}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}\right)−\mathrm{2}\:+\sqrt{\mathrm{4}{x}−\mathrm{1}}\left\{{arctan}\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{4}{x}−\mathrm{1}}}\right)−{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}{x}−\mathrm{1}}}\right)\right\} \\ $$
Commented by mathmax by abdo last updated on 10/Nov/19
2) ∫_0 ^1 ln(t^2 +t+(√2))dt =f((√2)) =(3/2)ln(2+(√2))−(1/4)ln(2)−2+(√(4(√2)−1)){ arctan((3/( (√(4(√2)−1)))))−arctan((1/( (√(4(√2)−1)))))}
$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({t}^{\mathrm{2}} \:+{t}+\sqrt{\mathrm{2}}\right){dt}\:={f}\left(\sqrt{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{2}\right)−\mathrm{2}+\sqrt{\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{1}}\left\{\:{arctan}\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{1}}}\right)−{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{1}}}\right)\right\} \\ $$
Answered by mind is power last updated on 10/Nov/19
f(x)=∫_0 ^1 ln(x+t+t^2 )dt=ln(x+2)−ln(x)−∫(((1+2t)t)/(x+t+t^2 ))dt {by part} =ln(((x+2)/2))−∫_0 ^1 ((2t^2 +2t+2x−t−(1/2)+(1/2)−2x)/(t^2 +t+x))dt =ln(((x+2)/x))−2∫1dt+∫((t+(1/2))/(t^2 +t+x))+(2x−(1/2))∫_0 ^1 (dt/((t+(1/2))^2 +((4x−1)/4))) =ln(((x+2)/x))−2+(1/2)ln{(((x+2)/x))}+∫_0 ^1 (dt/((((2t)/( (√(4x−1))))+(1/( (√(4x−1)))))^2 +1)) =(3/2)ln(((x+2)/x))−2+((√(4x−1))/2).[arctan(((2t+1)/( (√(4x−1)))))]_0 ^1 =(3/2)ln(((x+2)/x))−2+((√(4x−1))/2).[arctab((3/( (√(4x−1)))))−arctan((1/( (√(4x−1)))))] 2)x=(√2)
$$\mathrm{f}\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{x}+\mathrm{t}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{dt}=\mathrm{ln}\left(\mathrm{x}+\mathrm{2}\right)−\mathrm{ln}\left(\mathrm{x}\right)−\int\frac{\left(\mathrm{1}+\mathrm{2t}\right)\mathrm{t}}{\mathrm{x}+\mathrm{t}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:\:\:\left\{\mathrm{by}\:\mathrm{part}\right\} \\ $$$$=\mathrm{ln}\left(\frac{\mathrm{x}+\mathrm{2}}{\mathrm{2}}\right)−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2t}^{\mathrm{2}} +\mathrm{2t}+\mathrm{2x}−\mathrm{t}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2x}}{\mathrm{t}^{\mathrm{2}} +\mathrm{t}+\mathrm{x}}\mathrm{dt} \\ $$$$=\mathrm{ln}\left(\frac{\mathrm{x}+\mathrm{2}}{\mathrm{x}}\right)−\mathrm{2}\int\mathrm{1dt}+\int\frac{\mathrm{t}+\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{t}^{\mathrm{2}} +\mathrm{t}+\mathrm{x}}+\left(\mathrm{2x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dt}}{\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{4x}−\mathrm{1}}{\mathrm{4}}} \\ $$$$=\mathrm{ln}\left(\frac{\mathrm{x}+\mathrm{2}}{\mathrm{x}}\right)−\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left\{\left(\frac{\mathrm{x}+\mathrm{2}}{\mathrm{x}}\right)\right\}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dt}}{\left(\frac{\mathrm{2t}}{\:\sqrt{\mathrm{4x}−\mathrm{1}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{4x}−\mathrm{1}}}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{x}+\mathrm{2}}{\mathrm{x}}\right)−\mathrm{2}+\frac{\sqrt{\mathrm{4x}−\mathrm{1}}}{\mathrm{2}}.\left[\mathrm{arctan}\left(\frac{\mathrm{2t}+\mathrm{1}}{\:\sqrt{\mathrm{4x}−\mathrm{1}}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{x}+\mathrm{2}}{\mathrm{x}}\right)−\mathrm{2}+\frac{\sqrt{\mathrm{4x}−\mathrm{1}}}{\mathrm{2}}.\left[\mathrm{arctab}\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{4x}−\mathrm{1}}}\right)−\mathrm{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{4x}−\mathrm{1}}}\right)\right] \\ $$$$\left.\mathrm{2}\right)\mathrm{x}=\sqrt{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 10/Nov/19
thanks sir.
$${thanks}\:{sir}. \\ $$
Commented by mind is power last updated on 10/Nov/19
y′re welcom
$$\mathrm{y}'\mathrm{re}\:\mathrm{welcom} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *