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 206557 by luciferit last updated on 18/Apr/24 Answered by Frix last updated on 18/Apr/24 $$\mathrm{Without}\:\mathrm{substitution}: \\ $$$$\int\frac{\mathrm{3}+\mathrm{2sin}^{\mathrm{2}} \:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}}{dx}=\int\left(\frac{\mathrm{5}}{\mathrm{cos}^{\mathrm{2}} \:{x}}−\mathrm{2}\right){dx}= \\ $$$$=\mathrm{5tan}\:{x}\:−\mathrm{2}{x}\:+{C} \\…
Question Number 206558 by luciferit last updated on 18/Apr/24 Answered by Frix last updated on 18/Apr/24 $$\int\sqrt{\mathrm{3}+\mathrm{5}\sqrt{{x}}}{dx}\:\overset{{t}=\sqrt{\mathrm{3}+\mathrm{5}\sqrt{{x}}}} {=}\: \\ $$$$=\frac{\mathrm{4}}{\mathrm{25}}\int\left({t}^{\mathrm{4}} −\mathrm{3}{t}^{\mathrm{2}} \right){dt}=\frac{\mathrm{4}}{\mathrm{25}}\left(\frac{{t}^{\mathrm{5}} }{\mathrm{5}}−{t}^{\mathrm{3}} \right)= \\…
Question Number 206554 by cortano21 last updated on 18/Apr/24 Answered by starsouf last updated on 18/Apr/24 $$ \\ $$ Answered by mr W last updated…
Question Number 206549 by Shrodinger last updated on 18/Apr/24 $$\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{−{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx} \\ $$ Commented by Frix last updated on 18/Apr/24…
Question Number 206550 by SANOGO last updated on 18/Apr/24 $${calcul}\:\:{Residus}\:{de}\:{f}\:{en}\:{o} \\ $$$${f}\left({z}\right)=\frac{{ze}^{{z}} }{\left({z}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Answered by Berbere last updated on 18/Apr/24 $${f}\:\:{n}'{a}\:{pas}\:{de}\:{poles}\:{en}\:\mathrm{0};{Res}\left({f},\mathrm{0}\right)=\mathrm{0} \\…
Question Number 206572 by mr W last updated on 18/Apr/24 Commented by mr W last updated on 18/Apr/24 $${radius}\:{of}\:{circles}\:{is}\:\mathrm{1}. \\ $$$${find}\:{AB}=? \\ $$ Answered by…
Question Number 206541 by luciferit last updated on 18/Apr/24 Answered by lepuissantcedricjunior last updated on 18/Apr/24 $$\int\frac{\mathrm{3}\boldsymbol{{x}}+\mathrm{5}}{\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{10}\boldsymbol{{x}}+\mathrm{51}}}\boldsymbol{{dx}}=\boldsymbol{{k}} \\ $$$$\boldsymbol{{k}}=\frac{\mathrm{3}}{\mathrm{2}}\int\frac{\mathrm{2}\boldsymbol{{x}}−\mathrm{5}+\frac{\mathrm{10}}{\mathrm{3}}+\mathrm{5}}{\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{10}\boldsymbol{{x}}+\mathrm{51}}}\boldsymbol{{dx}} \\ $$$$\boldsymbol{{k}}=\mathrm{3}\sqrt{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{10}\boldsymbol{{x}}+\mathrm{51}}+\frac{\mathrm{25}}{\mathrm{2}}\int\frac{\boldsymbol{{dx}}}{\:\sqrt{\left(\boldsymbol{{x}}−\mathrm{5}\right)^{\mathrm{2}} +\mathrm{26}}}…
Question Number 206542 by luciferit last updated on 18/Apr/24 Answered by lepuissantcedricjunior last updated on 18/Apr/24 $$\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}}…