Question Number 206542 by luciferit last updated on 18/Apr/24
Answered by lepuissantcedricjunior last updated on 18/Apr/24
$$\int\left(\mathrm{3}\boldsymbol{{x}}+\mathrm{5}\right)\boldsymbol{{arctan}}\left(\boldsymbol{{x}}\right)\boldsymbol{{dx}}=\boldsymbol{{k}} \\ $$$$\begin{cases}{\boldsymbol{{u}}=\boldsymbol{{arctan}}\left(\boldsymbol{{x}}\right)}\\{\boldsymbol{{v}}'=\left(\mathrm{3}\boldsymbol{{x}}+\mathrm{5}\right)}\end{cases}=>\begin{cases}{\boldsymbol{{u}}'=\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }}\\{\boldsymbol{{v}}=\frac{\mathrm{3}}{\mathrm{2}}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{5}\boldsymbol{{x}}}\end{cases} \\ $$$$\boldsymbol{{k}}=\left(\frac{\mathrm{3}}{\mathrm{2}}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{5}\boldsymbol{{x}}\right)\boldsymbol{{arctan}}\left(\boldsymbol{{x}}\right)−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{10}}{\mathrm{3}}\boldsymbol{{x}}−\mathrm{1}}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }\boldsymbol{{dx}} \\ $$$$\boldsymbol{{k}}=\left(\frac{\mathrm{3}}{\mathrm{2}}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{5}\boldsymbol{{x}}\right)\boldsymbol{{arctan}}\left(\boldsymbol{{x}}\right)−\frac{\mathrm{3}}{\mathrm{2}}\boldsymbol{{x}}\:−\frac{\mathrm{5}}{\mathrm{2}}\boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \right)+\frac{\mathrm{3}}{\mathrm{2}}\boldsymbol{{arctan}}\left(\boldsymbol{{x}}\right)+\boldsymbol{{c}} \\ $$$$\boldsymbol{{k}}=\frac{\mathrm{3}}{\mathrm{2}}\left(\boldsymbol{{x}}^{\mathrm{2}} +\frac{\mathrm{10}}{\mathrm{3}}\boldsymbol{{x}}+\mathrm{1}\right)\boldsymbol{{arctan}}\left(\boldsymbol{{x}}\right)−\frac{\mathrm{3}}{\mathrm{2}}\boldsymbol{{x}}−\frac{\mathrm{5}}{\mathrm{2}}\boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \right)+\boldsymbol{{c}} \\ $$