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Question Number 200257 by Calculusboy last updated on 16/Nov/23

Answered by witcher3 last updated on 16/Nov/23

$${\ln }^2\left(\mathrm{x}\right)+1=\mathrm{y}\Rightarrow \mathrm{dy}=2\frac{\ln \left(\mathrm{x}\right)}{\mathrm{x}}$$$$=\int {\mathrm{y}}^2{\tan }^{-1}\left(\mathrm{y}\right)\mathrm{dy}$$$$=\frac{{\mathrm{y}}^3}{3}{\tan }^{-1}\left(\mathrm{y}\right)-\frac{1}{3}\int \frac{{\mathrm{y}}^3}{1+{\mathrm{y}}^2}\mathrm{dy}$$$$=\frac{{\mathrm{y}}^3{\tan }^{-1}\left(\mathrm{y}\right)}{3}-\frac{{\mathrm{y}}^2}{6}+\frac{1}{6}\ln (1+{\mathrm{y}}^2)+\mathrm{c}\mid$$

Commented by Calculusboy last updated on 16/Nov/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

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