Question Number 108839 by 150505R last updated on 19/Aug/20 Commented by Her_Majesty last updated on 20/Aug/20 $${this}\:{looks}\:{to}\:{me}\:{as}\:{if}\:{the}\:{integral}\:{must}\:{be} \\ $$$$\frac{{p}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }{{x}^{{q}} }\:{with}\:{p},\:{q}\:\in\mathbb{Q} \\ $$$$\frac{{d}}{{dx}}\left[\frac{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}}…
Question Number 108821 by mathmax by abdo last updated on 19/Aug/20 $$\mathrm{find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{lnx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$ Answered by mathmax by abdo last updated…
Question Number 174341 by behi834171 last updated on 31/Jul/22 $$\:\:\:\:\mathrm{1}.\underset{\:\:\:\:\mathrm{1}} {\overset{\:\:\:\:\:\:\:\:\:\:\mathrm{2}} {\int}}\:\sqrt{\boldsymbol{{x}}+\sqrt{\boldsymbol{{x}}−\mathrm{1}}\:}\:\boldsymbol{{dx}}=? \\ $$$$\:\:\:\:\:\mathrm{2}.\:\:\:\underset{\:\:\:\:\:\mathrm{0}} {\overset{\:\:\:\:\:\:\:\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} {\int}}\:\:\frac{\mathrm{2}\boldsymbol{{sinx}}+\mathrm{4}\boldsymbol{{cosx}}}{\boldsymbol{{sinx}}+\boldsymbol{{cosx}}}\:\:\boldsymbol{{dx}}=? \\ $$ Commented by mnjuly1970 last updated on 30/Jul/22…
Question Number 174336 by cortano1 last updated on 30/Jul/22 $$\:\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}\:\frac{\mid{x}−\mathrm{2}\mid}{{x}^{\mathrm{2}} −\mathrm{4}{x}}\:{dx}\:=? \\ $$ Answered by Mathspace last updated on 30/Jul/22 $${I}=−\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}}…
Question Number 108789 by 150505R last updated on 19/Aug/20 Answered by 1549442205PVT last updated on 19/Aug/20 $$\mathrm{Choose}\:\mathrm{A}\:\mathrm{because}\:\int_{\mathrm{0}.\mathrm{5}} ^{\:\mathrm{1}} \frac{\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{4}\right)\mathrm{sinx}+\mathrm{x4cox}}\mathrm{dx} \\ $$$$=\mathrm{ln}\mid\mathrm{cosec}\:\mathrm{2y}+\mathrm{coty}\mid_{\mathrm{0}.\mathrm{5}} ^{\mathrm{1}} =−\mathrm{1}.\mathrm{97}……
Question Number 108786 by 150505R last updated on 19/Aug/20 Answered by mathmax by abdo last updated on 19/Aug/20 $$\mathrm{first}\:\mathrm{we}\:\mathrm{study}\:\mathrm{the}\:\mathrm{convergence}\:\mathrm{A}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{\mathrm{n}−\mathrm{1}\:} \mathrm{cos}\left(\mathrm{ax}\right)\mathrm{dx} \\ $$$$\mathrm{A}_{\mathrm{n}}…
Question Number 108761 by bemath last updated on 19/Aug/20 $$\:\:\:\frac{\vdots\frac{\mathcal{B}{e}}{\mathcal{M}{ath}}\vdots}{\bigstar} \\ $$$${If}\:\int_{−\mathrm{1}} ^{\:\:{a}} \:\frac{{x}+\mathrm{1}}{\left({x}+\mathrm{2}\right)^{\mathrm{4}} }\:=\:\frac{\mathrm{10}}{\mathrm{81}}\:,\:{then}\:{the}\:{value}\:{of} \\ $$$${a}−\mathrm{2}\:{is}\:\_\_\_ \\ $$ Answered by ajfour last updated on…
Question Number 108750 by mathmax by abdo last updated on 19/Aug/20 $$\mathrm{calculste}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}\mathrm{dx} \\ $$ Answered by mnjuly1970 last updated on 19/Aug/20 $$\:\mathrm{sol}….:\:\mathrm{put}\::\:{x}=\frac{\mathrm{1}}{{t}}\Rightarrow\Omega=\:\int_{\mathrm{0}}…
Question Number 108748 by john santu last updated on 18/Aug/20 Answered by bemath last updated on 19/Aug/20 $$\:\:\:\:{by}\:{parts} \\ $$$$\:\begin{cases}{{u}=\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)\:\Rightarrow{du}=\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\:{dx}}\\{{v}=\int\:\mathrm{sin}\:{x}\:{dx}=−\mathrm{cos}\:{x}}\end{cases} \\ $$$${J}=−\mathrm{cos}\:{x}\:\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)+\int\:\frac{\mathrm{cos}\:^{\mathrm{2}} {x}\:{dx}}{\mathrm{sin}\:{x}} \\ $$$${J}=−\mathrm{cos}\:{x}\:\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)+\int\:\frac{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}}…
Question Number 108749 by mathmax by abdo last updated on 19/Aug/20 $$\mathrm{calculste}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{4}} }\:\mathrm{dx} \\ $$ Answered by mnjuly1970 last updated on 19/Aug/20 $${f}\left({a}\right)=\int_{\mathrm{0}}…