Question Number 206639 by mathlove last updated on 21/Apr/24 Commented by Frix last updated on 21/Apr/24 $$=\underset{\frac{\pi}{\mathrm{12}}} {\overset{\frac{\pi}{\mathrm{11}}} {\int}}\left(\frac{\sqrt{{s}}}{\mathrm{2}\left(\sqrt{{c}}+\sqrt{{s}}\right)}−\frac{\sqrt{{cs}}}{{c}+\sqrt{{s}}\left(\sqrt{{c}}+\mathrm{1}\right)}+\frac{\left(\sqrt{{c}}−\mathrm{1}\right){s}+\sqrt{{cs}}}{\mathrm{2}\left(\left(\sqrt{{s}}+\mathrm{1}\right){c}+\left(\sqrt{{c}}+\mathrm{1}\right){s}\right)}\right){dx} \\ $$$$\mathrm{With}\:{c}=\mathrm{cos}\:{x}\:\wedge{s}=\mathrm{sin}\:{x} \\ $$$$\mathrm{I}\:\mathrm{doubt}\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{this}. \\ $$…
Question Number 206592 by NasaSara last updated on 19/Apr/24 Answered by namphamduc last updated on 20/Apr/24 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}\right)}{\mathrm{1}−{x}}{dx}=\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}\right)}{\mathrm{1}−{x}}{dx}−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}\right)}{\mathrm{1}−{x}}{dx} \\ $$$${x}\rightarrow{x}^{\mathrm{2}}…
Question Number 206579 by York12 last updated on 19/Apr/24 $$\mathrm{Evaluate}\::\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}}{d}\left({x}\right). \\ $$ Answered by Berbere last updated on 19/Apr/24 $$=\left[{ln}\left(\mathrm{1}+{x}\right){ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 206558 by luciferit last updated on 18/Apr/24 Answered by Frix last updated on 18/Apr/24 $$\int\sqrt{\mathrm{3}+\mathrm{5}\sqrt{{x}}}{dx}\:\overset{{t}=\sqrt{\mathrm{3}+\mathrm{5}\sqrt{{x}}}} {=}\: \\ $$$$=\frac{\mathrm{4}}{\mathrm{25}}\int\left({t}^{\mathrm{4}} −\mathrm{3}{t}^{\mathrm{2}} \right){dt}=\frac{\mathrm{4}}{\mathrm{25}}\left(\frac{{t}^{\mathrm{5}} }{\mathrm{5}}−{t}^{\mathrm{3}} \right)= \\…
Question Number 206557 by luciferit last updated on 18/Apr/24 Answered by Frix last updated on 18/Apr/24 $$\mathrm{Without}\:\mathrm{substitution}: \\ $$$$\int\frac{\mathrm{3}+\mathrm{2sin}^{\mathrm{2}} \:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}}{dx}=\int\left(\frac{\mathrm{5}}{\mathrm{cos}^{\mathrm{2}} \:{x}}−\mathrm{2}\right){dx}= \\ $$$$=\mathrm{5tan}\:{x}\:−\mathrm{2}{x}\:+{C} \\…
Question Number 206549 by Shrodinger last updated on 18/Apr/24 $$\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{−{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx} \\ $$ Commented by Frix last updated on 18/Apr/24…
Question Number 206541 by luciferit last updated on 18/Apr/24 Answered by lepuissantcedricjunior last updated on 18/Apr/24 $$\int\frac{\mathrm{3}\boldsymbol{{x}}+\mathrm{5}}{\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{10}\boldsymbol{{x}}+\mathrm{51}}}\boldsymbol{{dx}}=\boldsymbol{{k}} \\ $$$$\boldsymbol{{k}}=\frac{\mathrm{3}}{\mathrm{2}}\int\frac{\mathrm{2}\boldsymbol{{x}}−\mathrm{5}+\frac{\mathrm{10}}{\mathrm{3}}+\mathrm{5}}{\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{10}\boldsymbol{{x}}+\mathrm{51}}}\boldsymbol{{dx}} \\ $$$$\boldsymbol{{k}}=\mathrm{3}\sqrt{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{10}\boldsymbol{{x}}+\mathrm{51}}+\frac{\mathrm{25}}{\mathrm{2}}\int\frac{\boldsymbol{{dx}}}{\:\sqrt{\left(\boldsymbol{{x}}−\mathrm{5}\right)^{\mathrm{2}} +\mathrm{26}}}…
Question Number 206543 by luciferit last updated on 18/Apr/24 Answered by lepuissantcedricjunior last updated on 18/Apr/24 $$\int\frac{\mathrm{3}+\mathrm{2}\sqrt{\boldsymbol{{x}}}}{\mathrm{4}+\sqrt{\boldsymbol{{x}}}}\boldsymbol{{dx}}=\int\frac{\mathrm{2}\left(\mathrm{4}+\sqrt{{x}}\right)−\mathrm{5}}{\mathrm{4}+\sqrt{\boldsymbol{{x}}}}\boldsymbol{{dx}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\boldsymbol{{x}}−\left\{\mathrm{5}\int\frac{\boldsymbol{{dx}}}{\mathrm{4}+\sqrt{\boldsymbol{{x}}}}\:\:\:\:\boldsymbol{{x}}=\boldsymbol{{t}}^{\mathrm{2}} \Leftrightarrow\boldsymbol{{dx}}=\mathrm{2}\boldsymbol{{tdt}}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\boldsymbol{{x}}−\left\{\mathrm{10}\int\left(\mathrm{1}−\frac{\mathrm{4}}{\mathrm{4}+\boldsymbol{{t}}}\right)\boldsymbol{{dt}}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\boldsymbol{{x}}−\mathrm{10}\sqrt{\boldsymbol{{x}}}+\mathrm{40}\boldsymbol{{ln}}\left(\mathrm{4}+\sqrt{\boldsymbol{{x}}}\right)+\boldsymbol{{c}} \\…
Question Number 206542 by luciferit last updated on 18/Apr/24 Answered by lepuissantcedricjunior last updated on 18/Apr/24 $$\int\left(\mathrm{3}\boldsymbol{{x}}+\mathrm{5}\right)\boldsymbol{{arctan}}\left(\boldsymbol{{x}}\right)\boldsymbol{{dx}}=\boldsymbol{{k}} \\ $$$$\begin{cases}{\boldsymbol{{u}}=\boldsymbol{{arctan}}\left(\boldsymbol{{x}}\right)}\\{\boldsymbol{{v}}'=\left(\mathrm{3}\boldsymbol{{x}}+\mathrm{5}\right)}\end{cases}=>\begin{cases}{\boldsymbol{{u}}'=\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }}\\{\boldsymbol{{v}}=\frac{\mathrm{3}}{\mathrm{2}}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{5}\boldsymbol{{x}}}\end{cases} \\ $$$$\boldsymbol{{k}}=\left(\frac{\mathrm{3}}{\mathrm{2}}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{5}\boldsymbol{{x}}\right)\boldsymbol{{arctan}}\left(\boldsymbol{{x}}\right)−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{10}}{\mathrm{3}}\boldsymbol{{x}}−\mathrm{1}}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}}…
Question Number 206536 by cortano21 last updated on 18/Apr/24 $$\:\:\:\:\underline{\underbrace{\lessdot}\cancel{} }\underbrace{\nsupseteqq\spadesuit\left[}\mathrm{2}^{{x}} \:\mathrm{sin}\:\left(\sqrt{\mathrm{4}−\mathrm{2}^{{x}+\mathrm{2}} }\:\right)\right. \\ $$ Commented by Frix last updated on 18/Apr/24 $$\mathrm{Simply}\:\mathrm{use}\:{t}=\sqrt{\mathrm{1}−\mathrm{2}^{{x}} } \\…