Question Number 17009 by arnabpapu550@gmail.com last updated on 29/Jun/17 $$\int_{\mathrm{0}} ^{\:\mathrm{a}} \:\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$ Commented by prakash jain last updated on 29/Jun/17…
Question Number 82531 by Power last updated on 22/Feb/20 Commented by abdomathmax last updated on 24/Feb/20 $${I}\:=\int\:\frac{\sqrt{\mathrm{1}−{x}}}{{x}}{dx}\:+\int\:\frac{\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }}{{x}}{dx}\:+\int\:\:\frac{\sqrt{\mathrm{4}−{x}^{\mathrm{4}} }}{{x}}{dx}\:{we}\:{have} \\ $$$$\int\:\:\frac{\sqrt{\mathrm{1}−{x}}}{{x}}{dx}\:\:{changement}\:\sqrt{\mathrm{1}−{x}}={t}\:{give}\mathrm{1}−{x}={t}^{\mathrm{2}} \:\Rightarrow \\ $$$${x}=\mathrm{1}−{t}^{\mathrm{2}} \:\Rightarrow\int\:\:\frac{\sqrt{\mathrm{1}−{x}}}{{x}}{dx}\:=\:\int\:\:\frac{{t}}{\mathrm{1}−{t}^{\mathrm{2}}…
Question Number 82510 by jagoll last updated on 22/Feb/20 $$\int\:\sqrt{{x}+\sqrt{{x}}\:}\:{dx}\:=\:? \\ $$ Commented by mathmax by abdo last updated on 23/Feb/20 $${I}\:=\int\sqrt{{x}+\sqrt{{x}}}{dx}\:{changement}\:\sqrt{{x}}={t}\:{give}\:{x}={t}^{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\int\sqrt{{t}^{\mathrm{2}}…
Question Number 148026 by mnjuly1970 last updated on 25/Jul/21 $$ \\ $$$$\mathrm{A}\::=\int_{−\mathrm{1}} ^{\:\mathrm{0}} {e}^{\:{x}\:+\frac{\mathrm{1}}{{x}}} \:\:\&\:{a}\mathrm{A}+{e}^{{b}} =\:\int_{\mathrm{0}} ^{\:\infty} \left(\frac{\mathrm{1}}{{e}}\right)^{\:{x}+\frac{\mathrm{1}}{{x}}} \\ $$$$\:\:{than}\::\:\:{a}+\:{b}\:=?\:\:\:\:\:{a}\:,\:{b}\:\in\:\mathbb{Z} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$ Answered…
Question Number 148020 by qaz last updated on 25/Jul/21 $$\underset{\mathrm{k}=\mathrm{2}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \centerdot\frac{\mathrm{lnk}}{\mathrm{k}}=\gamma\mathrm{ln2}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \mathrm{2} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 148000 by vvvv last updated on 25/Jul/21 $$\underset{−\infty} {\overset{+\infty} {\int}}\frac{\boldsymbol{{dx}}}{\left(\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$ Answered by gsk2684 last updated on 25/Jul/21 $$\mathrm{2}\underset{{x}=\mathrm{0}}…
Question Number 82450 by M±th+et£s last updated on 21/Feb/20 $$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\mathrm{1}+\left({tan}\left({x}\right)\right)^{\sqrt{\mathrm{2}}} }\:{dx} \\ $$ Commented by MJS last updated on 21/Feb/20 $$\mathrm{tan}^{\sqrt{\mathrm{2}}} \:{x}\:\notin\mathbb{R}\:\mathrm{for}\:\frac{\pi}{\mathrm{2}}<{x}<\pi \\…
Question Number 82446 by M±th+et£s last updated on 21/Feb/20 $${show}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}^{−{log}\left({x}\right)} \:{x}\:{log}\left({x}\right)\:{dx}={e}\sqrt{\pi} \\ $$ Commented by abdomathmax last updated on 21/Feb/20 $${changement}\:{logx}={t}\:{give}\:{x}={e}^{{t}}…
Question Number 82442 by mathmax by abdo last updated on 21/Feb/20 $$\left.\mathrm{1}\right){find}\:\int\:\frac{\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx} \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx} \\ $$ Commented by mathmax…
Question Number 82440 by mathmax by abdo last updated on 21/Feb/20 $${find}\:\int\:\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}{dx} \\ $$ Commented by abdomathmax last updated on 24/Feb/20 $${let}\:{A}\:=\int\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}{dx}\:\:{changement}\:\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}={t} \\ $$$${hive}\:\:\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}={t}^{\mathrm{2}} \:\Rightarrow\mathrm{1}−{x}={t}^{\mathrm{2}}…