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Question-200257




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

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