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Question-201293

Question Number 201293 by sonukgindia last updated on 03/Dec/23 Answered by aleks041103 last updated on 03/Dec/23 $${I}=\int_{\mathrm{2}} ^{\:\infty} \frac{\mathrm{8}{arcsec}\left({x}/\mathrm{2}\right){dx}}{{x}^{\mathrm{3}} −\mathrm{4}{x}}= \\ $$$$=\int_{\mathrm{1}} ^{\:\infty} \frac{\mathrm{8}{arcsec}\left(\left(\mathrm{2}{x}\right)/\mathrm{2}\right)}{\left(\mathrm{2}{x}\right)^{\mathrm{3}} −\mathrm{4}\left(\mathrm{2}{x}\right)}{d}\left(\mathrm{2}{x}\right)=\mathrm{2}\int_{\mathrm{1}}…

Question-201290

Question Number 201290 by sonukgindia last updated on 03/Dec/23 Answered by Calculusboy last updated on 03/Dec/23 $$\boldsymbol{{Solution}}:\:\boldsymbol{{By}}\:\boldsymbol{{using}}\:\boldsymbol{{kings}}\:\boldsymbol{{rule}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\boldsymbol{{sin}}^{\boldsymbol{\varphi}} \left(\boldsymbol{{x}}\right)}{\boldsymbol{{sin}}^{\boldsymbol{\varphi}} \left(\boldsymbol{{x}}\right)+\boldsymbol{{cos}}^{\boldsymbol{\varphi}} \left(\boldsymbol{{x}}\right)}\boldsymbol{{dx}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}}…

Question-201291

Question Number 201291 by sonukgindia last updated on 03/Dec/23 Answered by aleks041103 last updated on 03/Dec/23 $${I}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right){dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${notice} \\ $$$${x}=\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}},\:{dx}=−\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}…

Question-201091

Question Number 201091 by MrGHK last updated on 29/Nov/23 Commented by Frix last updated on 29/Nov/23 $$\mathrm{Look}\:\mathrm{at}\:\mathrm{the}\:\mathrm{first}\:\mathrm{few}\:\mathrm{summands}: \\ $$$${i}=\mathrm{0}\:\rightarrow\:\mathrm{1} \\ $$$${i}=\mathrm{1}\:\rightarrow\:\frac{{n}}{\mathrm{2}{n}+\mathrm{4}} \\ $$$${i}=\mathrm{2}\:\rightarrow\:\frac{{n}^{\mathrm{2}} −{n}}{\mathrm{6}{n}+\mathrm{24}{n}+\mathrm{24}} \\…