Question and Answers Forum
All Questions Topic List
AllQuestion and Answers: Page 1487
Question Number 55282 Answers: 1 Comments: 1
$${calculatef}\left({a}\right)=\:\:\int\:\:\:\left(\mathrm{1}+\frac{{a}}{{x}^{\mathrm{2}} }\right){arctan}\left(\frac{{a}}{{x}}\right){dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{1}} ^{+\infty} \left(\mathrm{1}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\right){arctan}\left(\frac{\mathrm{2}}{{x}}\right){dx}\:. \\ $$
Question Number 55281 Answers: 0 Comments: 0
$${find}\:{the}\:{value}\:{of}\:\prod_{{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right) \\ $$
Question Number 55280 Answers: 1 Comments: 1
$${fint}\:{f}\left({t}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{tx}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx}\:. \\ $$
Question Number 55279 Answers: 0 Comments: 0
$${find}\:{L}\left(\:{e}^{−{x}} {ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right)\:\:\:{with}\:{L}\:{mean}\:{laplace} \\ $$$${transform} \\ $$
Question Number 55278 Answers: 0 Comments: 0
$${calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)....\left({x}+{n}\right)} \\ $$
Question Number 55276 Answers: 0 Comments: 1
$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{16}{n}^{\mathrm{2}} −\mathrm{1}} \\ $$
Question Number 55274 Answers: 0 Comments: 4
$${let}\:\varphi\left({a}\right)\:=\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\:{arctan}\left(\frac{{a}}{{x}}\right){dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\varphi\left({a}\right)\:{interms}\:{of}\:{a} \\ $$$$\left.\mathrm{2}\right)\:\:{calculate}\:\varphi^{'} \left({a}\right)\:{at}\:{form}\:{of}\:{integral}. \\ $$$$\left.\mathrm{3}\right)\:{determine}\:\varphi^{\left({n}\right)} \left({a}\right)\:\:{at}\:{form}\:{of}\:{integral}. \\ $$$$\left.\mathrm{4}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:{arctan}\left(\frac{\mathrm{2}}{{x}}\right){dx}\:. \\ $$
Question Number 55273 Answers: 0 Comments: 1
$${let}\:\:{f}\left({x}\right)\:=\int_{{x}^{\mathrm{2}} } ^{\mathrm{1}+{x}} \:\:\frac{{dt}}{\mathrm{1}+{t}+{t}^{\mathrm{2}} } \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({x}\right)\:{interms}\:{of}\:{x} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)\:{and}\:{lim}_{{x}\rightarrow+\infty} \:\:{f}\left({x}\right) \\ $$
Question Number 55271 Answers: 0 Comments: 0
$$\left.\mathrm{1}\right)\:{let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{cost}}{\mathrm{3}\:+{sin}\left({xt}\right)}{dt} \\ $$$${find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{tcos}\left({xt}\right){cost}}{\left(\mathrm{3}\:+{sin}\left({xt}\right)\right)^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{cost}}{\mathrm{3}+{sint}}\:\:{and}\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{t}\:{cos}^{\mathrm{2}} {t}}{\left(\mathrm{3}+{sint}\right)^{\mathrm{2}} }{dt} \\ $$
Question Number 55270 Answers: 0 Comments: 1
$${let}\:{f}\left({x}\right)={x}^{{n}} {arctan}\left({x}^{\mathrm{2}} \right)\:\:{with}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right){developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$$$ \\ $$
Question Number 55269 Answers: 0 Comments: 4
$${let}\:{f}\left({x}\right)\:=\left(\mathrm{2}{x}+\mathrm{1}\right){ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right){developp}\:{f}\:{at}\:{integr}\:{serie}. \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\:. \\ $$
Question Number 55268 Answers: 0 Comments: 1
$${calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}+\mathrm{3}} \\ $$
Question Number 55308 Answers: 1 Comments: 0
Question Number 55288 Answers: 2 Comments: 0
Question Number 55258 Answers: 0 Comments: 0
$$\mathrm{3}{x}+\mathrm{5}{y}=?_{} \\ $$
Question Number 55267 Answers: 0 Comments: 0
$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left(\mathrm{1}+{xtan}\theta\right){d}\theta \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{values}\:{of}\:{integrals}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left(\mathrm{1}+{tan}\theta\right)\:\:{and}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+\mathrm{2}{tan}\theta\right){d}\theta\:. \\ $$$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{tan}\theta}{\mathrm{1}+{xtan}\theta}\:{d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{\frac{{sin}\theta}{{cos}\theta}}{\mathrm{1}+{x}\frac{{sin}\theta}{{cos}\theta}}{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{sin}\theta}{{cos}\theta\:+{xsin}\theta}\:{d}\theta\:\:=_{{tan}\left(\frac{\theta}{\mathrm{2}}\right)={t}} \:\:\:\:\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\:\frac{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:+\frac{\mathrm{2}{xt}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\:\:\frac{\mathrm{4}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}−{t}^{\mathrm{2}} \:+\mathrm{2}{xt}\right)}{dt}\:=−\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\:\frac{\mathrm{4}{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} −\mathrm{2}{xt}\:−\mathrm{1}\right)}{dt}\:{let}\:{decompose} \\ $$$${F}\left({t}\right)\:=\:\frac{\mathrm{4}{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} −\mathrm{2}{xt}\:−\mathrm{1}\right)}\:\:{roots}\:{of}\:\:{t}^{\mathrm{2}} −\mathrm{2}{xt}\:−\mathrm{1} \\ $$$$\Delta^{'} ={x}^{\mathrm{2}} +\mathrm{1}\:\Rightarrow{t}_{\mathrm{1}} ={x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:{and}\:{t}_{\mathrm{2}} ={x}−\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left({t}\right)=\frac{{a}}{{t}−{t}_{\mathrm{1}} }\:+\frac{{b}}{{t}−{t}_{\mathrm{2}} }\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${a}\:={lim}_{{t}\rightarrow{t}_{\mathrm{1}} } \left({t}−{t}_{\mathrm{1}} \right){F}\left({t}\right)=\frac{\mathrm{4}{t}_{\mathrm{1}} }{\left({t}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{1}\right)\left({t}_{\mathrm{1}} −{t}_{\mathrm{2}} \right)}\:=\alpha \\ $$$${b}\:={lim}_{{t}\rightarrow{t}_{\mathrm{2}} } \left({t}−{t}_{\mathrm{2}} \right){F}\left({t}\right)\:=\frac{\mathrm{4}{t}_{\mathrm{2}} }{\left({t}_{\mathrm{2}} ^{\mathrm{2}} \:+\mathrm{1}\right)\left({t}_{\mathrm{2}} −{t}_{\mathrm{1}} \right)}\:=\beta\:\Rightarrow{F}\left({t}\right)=\frac{\alpha}{{t}−{t}_{\mathrm{1}} }\:+\frac{\beta}{{t}−{t}_{\mathrm{2}} }\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{0}=−\frac{\alpha}{{t}_{\mathrm{1}} }\:−\frac{\beta}{{t}_{\mathrm{2}} }\:+{d}\:\:\Rightarrow{d}\:=\frac{\alpha}{{t}_{\mathrm{1}} }\:+\frac{\beta}{{t}_{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)=\frac{\mathrm{2}}{−\mathrm{2}{x}}\:=−\frac{\mathrm{1}}{{x}}=\frac{\alpha}{\mathrm{1}−{t}_{\mathrm{1}} }\:+\frac{\beta}{\mathrm{1}−{t}_{\mathrm{2}} }\:+\frac{{c}+{d}}{\mathrm{2}}\:\Rightarrow\frac{\mathrm{1}}{{x}}\:=\frac{\alpha}{{t}_{\mathrm{1}} −\mathrm{1}}\:+\frac{\beta}{{t}_{\mathrm{2}} −\mathrm{1}}\:−\frac{{c}}{\mathrm{2}}\:−\frac{{d}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{c}}{\mathrm{2}}\:=\frac{\alpha}{{t}_{\mathrm{1}} −\mathrm{1}}\:+\frac{\beta}{{t}_{\mathrm{2}} −\mathrm{1}}\:−\frac{{d}}{\mathrm{2}}\:−\frac{\mathrm{1}}{{x}}\:\Rightarrow{c}\:=\frac{\mathrm{2}\alpha}{{t}_{\mathrm{1}} −\mathrm{1}}\:+\frac{\mathrm{2}\beta}{{t}_{\mathrm{2}} −\mathrm{1}}\:−{d}−\frac{\mathrm{2}}{{x}} \\ $$$$\int\:{F}\left({t}\right){dt}\:=\alpha{ln}\mid{t}−{t}_{\mathrm{1}} \mid\:+\beta{ln}\mid{t}−{t}_{\mathrm{2}} \mid\:+\frac{{c}}{\mathrm{2}}{ln}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\:+{d}\:{arctan}\left({t}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} {F}\left({t}\right){dt}\:=\left[\alpha{ln}\mid{t}−{t}_{\mathrm{1}} \mid+\beta{ln}\mid{t}−{t}_{\mathrm{2}} \mid\:+\frac{{c}}{\mathrm{2}}{ln}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \\ $$$$=\alpha{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{t}_{\mathrm{1}} \mid\:+\beta{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{t}_{\mathrm{2}} \mid\:+\frac{{c}}{\mathrm{2}}{ln}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)\: \\ $$$$\left.=\left.\alpha{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)+\beta{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:+\frac{{ln}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{2}}{c}\:={f}^{'} \left({x}\right)\:\Rightarrow \\ $$$$\left.{f}\left({x}\right)=\int\:\alpha{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\mid\right){dx}+\beta\int\:{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\mid{dx} \\ $$$$+\frac{{cx}}{\mathrm{2}}{ln}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)\:+{C}\:....{be}\:{continued}... \\ $$
Question Number 55296 Answers: 1 Comments: 0
$${simplify}\:{the}\:{following}\:{expression}\:{if} \\ $$$${x}<\mathrm{0} \\ $$$$\mid\mathrm{4}{x}−\sqrt{\left(\mathrm{3}{x}−\mathrm{1}\right)^{\mathrm{2}} }\mid \\ $$
Question Number 55295 Answers: 0 Comments: 0
Question Number 55245 Answers: 1 Comments: 0
Question Number 55240 Answers: 1 Comments: 0
$${make}\:{r}\:{the}\:{subject}\:{of}\:{the}\:{relation} \\ $$$${m}=\frac{\mathrm{4}\sqrt{{u}+{r}}}{{v}−{r}} \\ $$
Question Number 55237 Answers: 1 Comments: 1
$$\int_{\mathrm{0}} ^{\mathrm{3}} \int_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{2}} \right)\mathrm{dxdy} \\ $$
Question Number 55230 Answers: 0 Comments: 1
$${calculate}\:{lim}_{\xi\rightarrow\mathrm{0}} \:\:\:\:\:\int_{\mathrm{1}} ^{\mathrm{1}+\xi} \:\:\:\:\frac{{arctan}\left(\xi{t}\right)}{{t}}\:{dt}\:. \\ $$
Question Number 55229 Answers: 0 Comments: 1
$${calculate}\:{lim}_{{n}\rightarrow+\infty} \:\int_{\mathrm{0}} ^{{n}} \:\:\frac{{e}^{{nx}} }{\mathrm{1}+{nx}^{\mathrm{2}} }\:{dx}\:\:. \\ $$
Question Number 55224 Answers: 3 Comments: 0
$${log}_{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{7}{x}−\mathrm{2}\right)={log}_{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{6}\right)+{log}_{\mathrm{4}} \mathrm{8} \\ $$$${find}\:{x} \\ $$
Question Number 55223 Answers: 0 Comments: 5
Question Number 55217 Answers: 1 Comments: 0
Pg 1482 Pg 1483 Pg 1484 Pg 1485 Pg 1486 Pg 1487 Pg 1488 Pg 1489 Pg 1490 Pg 1491
Terms of Service
Privacy Policy
Contact: info@tinkutara.com