Question Number 26950 by Joel578 last updated on 31/Dec/17 $$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{6}{x}^{\mathrm{2}} \:−\:{x}\:−\:\mathrm{1}}{\mathrm{6}{x}^{\mathrm{2}} \:−\:\mathrm{5}{x}\:+\:\mathrm{1}}\:{dx} \\ $$ Commented by prakash jain last updated on 31/Dec/17 $$\mathrm{Integral}\:\mathrm{does}\:\mathrm{not}\:\mathrm{converge}.…
Question Number 26949 by Joel578 last updated on 31/Dec/17 $$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{xy}}\:{dx}\:{dy} \\ $$ Commented by abdo imad last updated on 31/Dec/17 $$\left.{I}=\:\int_{\mathrm{0}^{}…
Question Number 92484 by jagoll last updated on 07/May/20 $$\underset{\mathrm{5}} {\overset{\mathrm{7}} {\int}}\:\mathrm{x}^{\mathrm{2}} \:\mathrm{f}\:'''\left(\mathrm{x}\right)\:\mathrm{dx}\:=? \\ $$$$\mathrm{if}\:\mathrm{f}''\left(\mathrm{7}\right)\:=\:\mathrm{2}\:,\:\mathrm{f}''\left(\mathrm{5}\right)\:=\mathrm{1} \\ $$$$\mathrm{f}'\left(\mathrm{7}\right)\:=\:−\mathrm{2}\:,\:\mathrm{f}'\left(\mathrm{5}\right)\:=\:−\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{7}\right)=\:−\mathrm{3}\:,\:\mathrm{f}\left(\mathrm{5}\right)\:=\:−\mathrm{4}\: \\ $$$$ \\ $$ Answered by…
Question Number 158009 by zainaltanjung last updated on 30/Oct/21 $$ \\ $$$$ \\ $$ Answered by puissant last updated on 30/Oct/21 $$\Omega=\int_{\mathrm{2}} ^{\mathrm{4}} \frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}}{dx}=\left[{ln}\mid{x}^{\mathrm{2}}…
Question Number 92447 by mathmax by abdo last updated on 07/May/20 $${find}\:\int_{\frac{\mathrm{1}}{\mathrm{6}}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \:\:\frac{{dx}}{\:\sqrt{\mathrm{1}−\mathrm{3}{x}}+\sqrt{\mathrm{1}+\mathrm{3}{x}}} \\ $$ Commented by mathmax by abdo last updated on 07/May/20…
Question Number 157977 by tounghoungko last updated on 30/Oct/21 $$\int\:\frac{{dx}}{\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{{x}}\:\right)\sqrt{{x}}}\:=? \\ $$ Answered by puissant last updated on 30/Oct/21 $$\Omega=\int\frac{{dx}}{\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{{x}}\right)\sqrt{{x}}} \\ $$$${u}=\sqrt{{x}\:}\:\rightarrow\:{du}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}{dx}\:\rightarrow\:{dx}=\:\mathrm{2}\sqrt{{x}}{du} \\ $$$$\Rightarrow\:\Omega\:=\:\int\frac{\mathrm{2}\sqrt{{x}}{du}}{\left(\mathrm{1}+\sqrt{{u}}\right)\sqrt{{x}}}\:=\:\mathrm{2}\int\frac{{du}}{\left(\mathrm{1}+\sqrt{{u}}\right)} \\…
Question Number 157961 by mnjuly1970 last updated on 30/Oct/21 $$ \\ $$$$\:\:\:\:\:\:\:{prove}\:{that}: \\ $$$$ \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\:\infty} {x}^{\:\mathrm{2}} {tanh}\left({x}\right).{e}^{\:−{x}} {dx}=\frac{\pi^{\:\mathrm{3}} }{\mathrm{8}}\:−\mathrm{2}\:\:\:\:\:\:\: \\ $$$$ \\ $$…
Question Number 92420 by I want to learn more last updated on 06/May/20 Commented by mathmax by abdo last updated on 07/May/20 $${I}\:=\int\:\:\frac{{dx}}{{cos}^{\mathrm{6}} {x}\:+{sin}^{\mathrm{6}} {x}}\:\Rightarrow\:{I}\:=\int\:\:\:\frac{{dx}}{\left({cos}^{\mathrm{2}}…
Question Number 92421 by mhmd last updated on 06/May/20 $$\int\frac{{cscx}}{{cos}\left(\mathrm{2}{x}\right)+\mathrm{2}{cos}^{\mathrm{2}} {x}}{dx} \\ $$$${pleas}\:{sir}\:{help}\:{me} \\ $$ Answered by niroj last updated on 07/May/20 $$\:\:\int\:\:\frac{\:\:\mathrm{cosec}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{2x}+\:\mathrm{2cos}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}\: \\…
Question Number 92410 by mathmax by abdo last updated on 06/May/20 $${find}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}+\mathrm{3}{x}}−\sqrt{\mathrm{1}−\mathrm{3}{x}}} \\ $$ Commented by Prithwish Sen 1 last updated on 06/May/20…