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Category: Integration

Question-201509

Question Number 201509 by Calculusboy last updated on 07/Dec/23 Answered by witcher3 last updated on 09/Dec/23 $$\mathrm{tack}\:\mathrm{principal}\:\:\mathrm{definition}\:\mathrm{of}\:\mathrm{Log}\left(\mathrm{z}\right)=\mathrm{ln}\mid\mathrm{z}\mid+\mathrm{iarg}\left(\mathrm{z}\right) \\ $$$$\left.\mathrm{z}\in\right]−\frac{\pi}{\mathrm{2}},\frac{\mathrm{3}\pi}{\mathrm{2}}\left[\:\right. \\ $$$$\mathrm{ln}\left(\mathrm{ix}+\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)=\mathrm{ln}\mid\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{ln}^{\mathrm{2}} \left(\mathrm{cos}\left(\mathrm{x}\right)\right.}\mid+\mathrm{i}\:\mathrm{arg}\left(\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{x}\right)+\mathrm{ix}\right)\right.\right. \\ $$$$\mathrm{arg}\left(\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{x}\right)+\mathrm{ix}\right)\in\left[\frac{\pi}{\mathrm{2}},\pi\left[\:\right.\right.\right.…

Question-201452

Question Number 201452 by tri26112004 last updated on 06/Dec/23 Answered by Calculusboy last updated on 06/Dec/23 $$\int\boldsymbol{{x}}^{−\mathrm{2}} \boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} \boldsymbol{{dx}} \\ $$$$\boldsymbol{{Solution}}:\:\:\boldsymbol{{by}}\:\boldsymbol{{using}}\:\boldsymbol{{IBP}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{u}}=\boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} \:\:\:\boldsymbol{{du}}=−\mathrm{4}\boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} \boldsymbol{{dx}}\:\:\boldsymbol{{dv}}=\boldsymbol{{x}}^{−\mathrm{2}}…