Question Number 63782 by mathmax by abdo last updated on 09/Jul/19 $${let}\:{f}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:\:{with}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right){calculste}\:{also}\:{g}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{4}}…
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Question Number 129277 by liberty last updated on 14/Jan/21 $$\:\mathrm{O}\:=\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{arctan}\:\left(\frac{\mathrm{3X}+\mathrm{3}}{\mathrm{1}−\mathrm{2X}−\mathrm{X}^{\mathrm{2}} }\right)}{\mathrm{1}+\mathrm{X}^{\mathrm{2}} }\:\mathrm{dX} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 129274 by bramlexs22 last updated on 14/Jan/21 $$\:\int\:\frac{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:\mathrm{x}}{\mathrm{1}+\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}\:\mathrm{dx}\:? \\ $$ Answered by liberty last updated on 14/Jan/21 $$\:\int\:\frac{\mathrm{cos}\:\mathrm{x}\left(\mathrm{cos}\:\mathrm{x}+\mathrm{1}\right)}{\mathrm{sin}\:\mathrm{x}+\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}\:\mathrm{dx}\:=\:\int\:\frac{\mathrm{cos}\:\mathrm{x}\left(\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)}{\mathrm{2cos}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\left[\:\mathrm{sin}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right]}\:\mathrm{dx}= \\ $$$$\:\int\:\frac{\mathrm{cos}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\left(\mathrm{cos}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)+\mathrm{sin}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)\left(\mathrm{cos}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{sin}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)}{\mathrm{sin}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}\mathrm{dx}=…
Question Number 63738 by MJS last updated on 08/Jul/19 $$\mathrm{useful}\:\mathrm{formula} \\ $$$$======== \\ $$$$ \\ $$$$\forall{a}\in\mathbb{R}^{+} :\forall{b}\:\in\mathbb{R}:\:{a}\:\mathrm{sin}\:{x}\:+{b}\:\mathrm{cos}\:{x}\:=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\mathrm{sin}\:\left({x}+\mathrm{arctan}\:\frac{{b}}{{a}}\right) \\ $$$$\int\frac{{dx}}{{a}\:\mathrm{sin}\:{x}\:+{b}\:\mathrm{cos}\:{x}}= \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\int\frac{{dx}}{\mathrm{sin}\:\left({x}+\mathrm{arctan}\:\frac{{b}}{{a}}\right)}=…
Question Number 129275 by abdurehime last updated on 14/Jan/21 $$\int\mathrm{e}^{\mathrm{x}^{\mathrm{x}} } \mathrm{dx}?????? \\ $$ Commented by abdurehime last updated on 14/Jan/21 $$\mathrm{no}\:\mathrm{one}???? \\ $$ Commented…
Question Number 129273 by abdurehime last updated on 14/Jan/21 $$\mathrm{please}\:\mathrm{answer}\:\mathrm{my}\:\mathrm{question} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 129269 by bramlexs22 last updated on 14/Jan/21 $$\:\int_{\mathrm{1}} ^{{a}} \:\frac{\mathrm{4}\sqrt{{x}}\:+{k}}{\:\sqrt{{x}}\:+\mathrm{1}}\:=\:\mathrm{4}{a}+\mathrm{3}\:.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:\int_{\mathrm{1}} ^{\:{a}} \frac{\mathrm{1}}{\:\sqrt{{x}}\:+\mathrm{1}}\:{dx}\:. \\ $$ Answered by liberty last updated on 14/Jan/21…
Question Number 63721 by mathmax by abdo last updated on 08/Jul/19 $${calculate}\:\int\:\:\frac{{dx}}{\:\sqrt{\left({x}−\mathrm{1}\right)\left(\mathrm{2}−{x}\right)}} \\ $$ Commented by Prithwish sen last updated on 08/Jul/19 $$\mathrm{put}\:\mathrm{x}−\mathrm{l}\:=\:\mathrm{z}^{\mathrm{2}} \:\:\:\Rightarrow\mathrm{dx}\:=\:\mathrm{2zdz} \\…
Question Number 63722 by mathmax by abdo last updated on 08/Jul/19 $$\left.\mathrm{1}\right)\:{calculate}\:\int\:\left({x}^{\mathrm{2}} −{x}+\mathrm{2}\right)\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx} \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{\mathrm{2}} −{x}+\mathrm{2}\right)\sqrt{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}}{dx}\:. \\ $$ Commented by Prithwish…