Question Number 200257 by Calculusboy last updated on 16/Nov/23
Answered by witcher3 last updated on 16/Nov/23
$$\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{1}=\mathrm{y}\Rightarrow\mathrm{dy}=\mathrm{2}\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{x}} \\ $$$$=\int\mathrm{y}^{\mathrm{2}} \mathrm{tan}^{−\mathrm{1}} \left(\mathrm{y}\right)\mathrm{dy} \\ $$$$=\frac{\mathrm{y}^{\mathrm{3}} }{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{y}\right)−\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{y}^{\mathrm{3}} }{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\mathrm{dy} \\ $$$$=\frac{\mathrm{y}^{\mathrm{3}} \mathrm{tan}^{−\mathrm{1}} \left(\mathrm{y}\right)}{\mathrm{3}}−\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)+\mathrm{c}\mid \\ $$
Commented by Calculusboy last updated on 16/Nov/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$