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Category: Integration

0-1-ln-2-1-x-lnx-x-dx-

Question Number 160281 by amin96 last updated on 27/Nov/21 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{lnx}}}{\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}}=? \\ $$ Answered by TheSupreme last updated on 27/Nov/21 $$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}^{\mathrm{2}}…

0-1-ln-1-x-ln-x-1-x-dx-

Question Number 160276 by amin96 last updated on 27/Nov/21 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{ln}}\left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)}{\mathrm{1}−\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}}=? \\ $$ Answered by mnjuly1970 last updated on 27/Nov/21 $$\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}\left({x}\right).{ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx}=\:\left[−{li}_{\mathrm{2}} \left({x}\right).{ln}\left({x}\right)\right]_{\mathrm{0}}…

2t-1-t-4-1-t-dt-

Question Number 94735 by student work last updated on 20/May/20 $$\int\frac{\mathrm{2t}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{4}} \right)\left(\mathrm{1}+\mathrm{t}\right)}\mathrm{dt}=? \\ $$ Commented by student work last updated on 20/May/20 $$\mathrm{who}\:\mathrm{can}\:\mathrm{another}? \\ $$…

tan-x-dx-tan-x-cot-x-2-dx-tan-x-cot-x-2-dx-1-2-sin-x-cos-x-sin-2x-dx-1-2-sin-x-cos-x-sin-2x-dx-1-2-sin-x-c

Question Number 94718 by john santu last updated on 20/May/20 $$\int\:\sqrt{\mathrm{tan}\:\mathrm{x}}\:\mathrm{dx}\:= \\ $$$$\int\frac{\sqrt{\mathrm{tan}\:\mathrm{x}}+\sqrt{\mathrm{cot}\:\mathrm{x}}}{\mathrm{2}}\:\mathrm{dx}\:+\:\int\:\frac{\sqrt{\mathrm{tan}\:\mathrm{x}}−\sqrt{\mathrm{cot}\:\mathrm{x}}}{\mathrm{2}}\:\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\int\:\frac{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}{\:\sqrt{\mathrm{sin}\:\mathrm{2x}}}\:\mathrm{dx}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\int\:\frac{\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}}{\:\sqrt{\mathrm{sin}\:\mathrm{2x}}}\:\mathrm{dx} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\int\:\frac{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}{\:\sqrt{\mathrm{1}−\left(\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} }}\:\mathrm{dx}\:+\: \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int\:\frac{\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}}{\:\sqrt{\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} −\mathrm{1}}}\:\mathrm{dx}\: \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\int\:\frac{\mathrm{dt}}{\:\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\int\:\frac{−\mathrm{du}}{\:\sqrt{\mathrm{u}^{\mathrm{2}} −\mathrm{1}}}…

tan-x-cot-x-dx-

Question Number 94662 by i jagooll last updated on 20/May/20 $$\int\:\sqrt{\mathrm{tan}\:\mathrm{x}+\mathrm{cot}\:\mathrm{x}}\:\mathrm{dx}\:=\:? \\ $$ Commented by i jagooll last updated on 20/May/20 $$\int\:\frac{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}}{\:\sqrt{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}\:.\sqrt{\mathrm{tan}\:\mathrm{x}}}\:=\: \\…