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Author: Tinku Tara

Question-200257

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}}…

Question-200159

Question Number 200159 by Calculusboy last updated on 15/Nov/23 Commented by 0670322918 last updated on 15/Nov/23 $$\underset{\mathrm{0}} {\int}^{\mathrm{1}} \frac{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}{dx}= \\…

Question-200186

Question Number 200186 by Calculusboy last updated on 15/Nov/23 Answered by ajfour last updated on 15/Nov/23 $$\frac{{y}}{{x}}={p}\:,\:\frac{{z}}{{y}}={q}\:,\:\frac{{x}}{{z}}={r}\:=\frac{\mathrm{1}}{{pq}} \\ $$$$\mathrm{1}+{p}+{p}^{\mathrm{2}} =\left(\mathrm{2}+{p}\right)\left(\frac{\mathrm{1}}{{r}}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\mathrm{1}+{q}+{q}^{\mathrm{2}} =\left(\mathrm{2}+{q}\right)\left(\frac{\mathrm{1}}{{p}}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\mathrm{1}+{r}+{r}^{\mathrm{2}}…