Question Number 131637 by rs4089 last updated on 07/Feb/21 Answered by Dwaipayan Shikari last updated on 07/Feb/21 $$\int_{−\infty} ^{\infty} \frac{{sinx}}{{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}=\int_{−\infty} ^{\infty} \frac{{sinx}}{{x}}−\frac{{xsinx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}=\pi−\frac{\pi}{{e}} \\…
Question Number 66096 by sitangshu17 last updated on 09/Aug/19 $$\int\:\:\frac{\sqrt{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}}{\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=\:? \\ $$ Commented by mathmax by abdo last updated on 09/Aug/19 $${let}\:{I}\:=\int\:\:\:\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}}…
Question Number 131630 by Ahmed1hamouda last updated on 07/Feb/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 66085 by mathmax by abdo last updated on 09/Aug/19 $${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{arctan}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}}}{dx} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 66082 by aliesam last updated on 08/Aug/19 Commented by mathmax by abdo last updated on 09/Aug/19 $${let}\:{I}\:=\int_{−\infty} ^{+\infty} \:{xsin}\left({x}^{\mathrm{3}} \right){dx}\:\Rightarrow{I}\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:{x}\:{sin}\left({x}^{\mathrm{3}} \right){dx}…
Question Number 131614 by mnjuly1970 last updated on 06/Feb/21 $$\:\:\:\:\:\:\:\:\:…{nice}\:\:{calculus}… \\ $$$$\:\:\:\:\:\:\:{prove}\:{that}\::\: \\ $$$$\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){cos}\left(\mathrm{2}\pi{nx}\right){dx}=\frac{\mathrm{1}}{\mathrm{4}{n}} \\ $$$${for}\:{example}\::\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){cos}\left(\mathrm{2}\pi{x}\right){dx}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Terms of Service…
Question Number 536 by kth last updated on 25/Jan/15 $$\int_{\mathrm{1}} ^{\mathrm{4}} \frac{\mathrm{2}}{\:\sqrt{{x}}\left(\sqrt{{x}}+\mathrm{4}\right)^{\mathrm{3}} }{dx} \\ $$$$ \\ $$ Answered by Vishal last updated on 24/Jan/15 $${let}\:\sqrt{{x}}+\mathrm{4}={t}\:{then}\:\frac{\mathrm{1}}{\:\sqrt{{x}}}\:{dx}=\mathrm{2}\:{dt}…
Question Number 537 by kth last updated on 25/Jan/15 $$\int_{\mathrm{4}} ^{\mathrm{5}} \frac{{x}−\mathrm{2}}{\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}\right)^{\mathrm{3}} }{dx} \\ $$ Answered by Vishal last updated on 25/Jan/15 $${let}\:{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}\:=\:{t}\:{then}\:\left({x}−\mathrm{2}\right)\:{dx}\:=\:\frac{{dt}}{\mathrm{2}}…
Question Number 535 by kth last updated on 25/Jan/15 $$\int_{\mathrm{4}} ^{\mathrm{1}} \frac{\mathrm{2}}{\:\sqrt{{x}}} \\ $$ Answered by 13/NaSaNa(N)056565 last updated on 25/Jan/15 $$=\mathrm{2}\int_{\mathrm{4}} ^{\mathrm{1}} {x}^{\frac{−\mathrm{1}}{\mathrm{2}}} {dx}…
Question Number 66064 by mathmax by abdo last updated on 08/Aug/19 $${find}\:{the}\:{value}\:{of}\:\int_{−\infty} ^{+\infty} \:{cos}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right){dx} \\ $$ Commented by mathmax by abdo last updated on…