Question Number 217441 by mr W last updated on 13/Mar/25 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{3}} \:\left(\mathrm{1}−{x}\right)}{{x}^{\mathrm{3}} }\:{dx}=? \\ $$ Answered by MrGaster last updated on 14/Mar/25 $$\int_{\mathrm{0}}…
Question Number 217356 by SciMaths last updated on 11/Mar/25 Answered by profcedricjunior last updated on 11/Mar/25 $$\boldsymbol{{i}}=\int_{\mathrm{1}} ^{\mathrm{2}} \int_{\boldsymbol{{y}}} ^{\boldsymbol{{y}}^{\mathrm{2}} } \int_{\mathrm{0}} ^{\boldsymbol{{ln}}\left(\boldsymbol{{y}}+\boldsymbol{{z}}\right)} \boldsymbol{{e}}^{\boldsymbol{{x}}} \boldsymbol{{dxdydz}}…
Question Number 217289 by Frix last updated on 08/Mar/25 $$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{sin}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}−{x}^{\mathrm{2}} }}\:{dx}=? \\ $$ Commented by Ghisom last updated on 11/Mar/25 $$\mathrm{we}\:\mathrm{can}\:\mathrm{use}\:\mathrm{partial}\:\mathrm{integration} \\…
Question Number 217290 by mnjuly1970 last updated on 08/Mar/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}^{\mathrm{2}} }\:{dx}=\:? \\ $$$$ \\ $$$$ \\ $$ Answered by profcedricjunior…
Question Number 217255 by MrGaster last updated on 07/Mar/25 $$\int_{−\infty} ^{+\infty} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx}=\sqrt{\mathrm{2}\pi},\int_{−\infty} ^{+\infty} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{x}} {dx}. \\ $$ Answered by vnm last updated…
Question Number 217211 by Ghisom last updated on 06/Mar/25 $$\mathrm{a}\:\mathrm{nice}\:\mathrm{one}: \\ $$$$\mathrm{prove}\:\:\:\:\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{−\frac{\mathrm{ln}\:{t}}{{t}}}\:{dt}=\sqrt{\mathrm{2}\pi} \\ $$ Answered by MrGaster last updated on 06/Mar/25 $$\left(\mathrm{1}\right):\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 217219 by mnjuly1970 last updated on 06/Mar/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{calculate} \\ $$$$\begin{array}{|c|}{\:\:\mathscr{L}\:\:\left(\:\int_{\mathrm{1}} ^{\:\infty} \frac{\:\mathrm{e}^{\:−{tx}} }{{x}}{dx}\:\right)\:\underset{\mathrm{transfom}} {\overset{\mathrm{laplace}} {=}}?\:\:;\:\:{t}>\mathrm{0}\:}\\\hline\end{array} \\ $$$$ \\ $$$$\:\:\: \\ $$…
Question Number 217122 by efronzo1 last updated on 01/Mar/25 $$\:\:\:\:\:\int\:\frac{\sqrt{\mathrm{cos}\:\mathrm{2x}}}{\mathrm{cos}\:\mathrm{x}}\:\mathrm{dx}\:=? \\ $$ Answered by Frix last updated on 01/Mar/25 $$\int\frac{\sqrt{\mathrm{cos}\:\mathrm{2}{x}}}{\mathrm{cos}\:{x}}{dx}=\int\frac{\sqrt{−\mathrm{1}+\mathrm{2cos}^{\mathrm{2}} \:{x}}}{\mathrm{cos}\:{x}}{dx}\:\overset{\left[{t}=\sqrt{\mathrm{2}}\mathrm{sin}\:{x}\right]} {=} \\ $$$$=\sqrt{\mathrm{2}}\int\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{\mathrm{2}−{t}^{\mathrm{2}}…
Question Number 216990 by MathematicalUser2357 last updated on 26/Feb/25 $$\left(\mathrm{1}\right)\:\int\left(\mathrm{sec}^{\mathrm{2}} {x}\centerdot\sqrt{\mathrm{tan}\:{x}}\right){dx}=? \\ $$ Commented by MathematicalUser2357 last updated on 26/Feb/25 $$\left(\mathrm{2}\right)\:\mathrm{Go}\:\mathrm{to}\:\underline{\mathrm{https}://\mathrm{playentry}.\mathrm{org}/\mathrm{project}/\mathrm{67a8a3c0d78abe99420525f0}} \\ $$$$\mathrm{and}\:\mathrm{press}\:\mathrm{the}\:\mathrm{play}\:\mathrm{button}\:\mathrm{and}\:\mathrm{screenshot}\:\mathrm{and}\:\mathrm{give}\:\mathrm{the}\:\mathrm{screenshot}\:\mathrm{to}\:\mathrm{me} \\ $$…
Question Number 216886 by Engr_Jidda last updated on 23/Feb/25 $${Evaluate}\:\frac{\underset{{k}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\left(\int_{\mathrm{0}} ^{{k}} \left(\mathrm{4}{u}+\mathrm{1}\right){du}\right)}{\mathrm{5}^{\mathrm{2}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}}\left(\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{m}^{\mathrm{2}} +\mathrm{2}{m}}\right)^{{n}−\mathrm{1}} }\int_{{sin}^{−\mathrm{1}} \left(\frac{−\sqrt{\mathrm{2}}}{\mathrm{2}}\right)} ^{\frac{\pi}{\mathrm{2}}{cos}\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{1}−{sec}\theta{sin}\theta}{\frac{{tan}\theta+{cot}\theta}{\varrho^{\theta} −\varrho^{\pi{i}}…