Menu Close

Category: Integration

find-the-value-of-n-1-1-n-1-H-2n-n-where-H-n-1-1-2-1-3-1-n-

Question Number 197819 by mnjuly1970 last updated on 30/Sep/23 $$ \\ $$$$\:\:\:\:\:\:{find}\:{the}\:{value}\:{of}\:\:: \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:{H}_{\:\mathrm{2}{n}} }{{n}}\:=\:? \\ $$$${where},{H}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+…+\frac{\mathrm{1}}{{n}} \\ $$ Answered by…

I-2-6-x-1-x-1-dx-

Question Number 197802 by cortano12 last updated on 29/Sep/23 $$\:\:\:\mathrm{I}=\underset{−\mathrm{2}} {\overset{\mathrm{6}} {\int}}\:\frac{\mid\mathrm{x}−\mathrm{1}\mid}{\mathrm{x}−\mathrm{1}}\:\mathrm{dx}\:=? \\ $$ Answered by MM42 last updated on 29/Sep/23 $${I}=\int_{−\mathrm{2}} ^{\mathrm{1}} −{dx}+\int_{\mathrm{1}} ^{\mathrm{6}}…

Question-197767

Question Number 197767 by AR19 last updated on 28/Sep/23 Commented by Frix last updated on 28/Sep/23 $$\int\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}{dx}={x}\:_{\mathrm{2}} {F}_{\mathrm{1}} \:\left(−\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{\mathrm{3}};\:\frac{\mathrm{4}}{\mathrm{3}};\:−{x}^{\mathrm{3}} \right)\:+{C} \\ $$ Terms of…

2-0-1-tan-1-x-dx-

Question Number 197744 by mathlove last updated on 27/Sep/23 $$\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} {x}\:{dx}=? \\ $$ Answered by witcher3 last updated on 27/Sep/23 $$\int\mathrm{2tan}^{−\mathrm{1}} \left(\mathrm{x}\right)\mathrm{dx}=\mathrm{2xtan}^{−\mathrm{1}} \left(\mathrm{x}\right)−\int\frac{\mathrm{2x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}}…

0-pi-2-dx-3-tan-x-

Question Number 197436 by horsebrand11 last updated on 17/Sep/23 $$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\mathrm{dx}}{\mathrm{3}+\mathrm{tan}\:\mathrm{x}}\:=? \\ $$ Answered by Frix last updated on 17/Sep/23 $$\int\frac{{dx}}{\mathrm{3}+\mathrm{tan}\:{x}}\:\overset{{t}=\mathrm{tan}\:{x}} {=}\:\int\frac{{dt}}{\left({t}+\mathrm{3}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}= \\…