Question Number 175051 by Mathspace last updated on 17/Aug/22 $${find}\:{the}\:{value}\:{of}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{arctanx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dx} \\ $$ Commented by mokys last updated on 20/Aug/22…
Question Number 109509 by bemath last updated on 24/Aug/20 $$\:\:\frac{{bemath}}{\underset{{i}={cooll}} {\overset{{nice}} {\sum}}\left({joss}\right)_{{i}} }\: \\ $$$$ \\ $$$$\int\:\frac{{x}^{\mathrm{2}} \:{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}} \\ $$ Answered by Dwaipayan Shikari…
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Question Number 175042 by peter frank last updated on 17/Aug/22 Answered by Frix last updated on 18/Aug/22 $$\int\frac{{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:{x}\:+{b}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:{x}}{{a}^{\mathrm{4}} \mathrm{sin}^{\mathrm{4}} \:{x}\:+{b}^{\mathrm{4}} \mathrm{cos}^{\mathrm{4}}…
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Question Number 43939 by abdo.msup.com last updated on 18/Sep/18 $${find}\:{f}\left(\xi\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\mathrm{1}+\left({t}−{i}\xi\right)^{\mathrm{2}} } \\ $$$${and}\:{calculate}\:{f}^{'} \left(\xi\right) \\ $$ Commented by maxmathsup by imad last updated…
Question Number 43938 by abdo.msup.com last updated on 18/Sep/18 $${calvulste}\:{A}_{{n}} =\int_{\mathrm{0}} ^{{n}} \:{t}^{\mathrm{2}} \left[\frac{\mathrm{1}}{\left({t}+\mathrm{1}\right)^{\mathrm{3}} }\right]{dt} \\ $$$${and}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$ Commented by maxmathsup by imad…
Question Number 43937 by abdo.msup.com last updated on 18/Sep/18 $${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5}}}{dx} \\ $$ Commented by maxmathsup by imad last updated on 19/Sep/18 $${let}\:{A}\:=\:\int_{\mathrm{0}}…
Question Number 109472 by john santu last updated on 24/Aug/20 Answered by 1549442205PVT last updated on 24/Aug/20 $$\mathrm{Put}\:\mathrm{F}=\int\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }} \\ $$$$\mathrm{Putting}\:\:\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }+\mathrm{1}=\mathrm{u}^{\mathrm{2}} \Rightarrow\mathrm{2udu}=\frac{−\mathrm{2}}{\mathrm{x}^{\mathrm{3}} }\mathrm{dx}…