Question and Answers Forum

All Questions   Topic List

AllQuestion and Answers: Page 651

Question Number 154202    Answers: 1   Comments: 0

𝛀 =∫_( 0) ^( ∞) ((lnβˆ™(1 + x))/(xβˆ™(x^2 + x + 1))) dx = ?

$$\boldsymbol{\Omega}\:\:=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{ln}\centerdot\left(\mathrm{1}\:+\:\mathrm{x}\right)}{\mathrm{x}\centerdot\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{1}\right)}\:\mathrm{dx}\:=\:? \\ $$

Question Number 154194    Answers: 0   Comments: 1

Question Number 154195    Answers: 3   Comments: 0

∫((x+sinx)/(1+cosx))dx please,help me

$$\int\frac{{x}+{sinx}}{\mathrm{1}+{cosx}}{dx}\:\:\:\: \\ $$$${please},{help}\:{me} \\ $$

Question Number 154192    Answers: 1   Comments: 0

Ξ© :=∫_0 ^( 1) (( x.sin(ln(x)))/(1βˆ’x))dx method 1 Ξ©= Im[∫_0 ^( 1) (( x^( i+1) )/(1βˆ’x)) dx=Ξ¦] Ξ¦ = ∫_0 ^( 1) (( x^( i+1) +x^( i+2) )/(1βˆ’x^( 2) ))dx =^(x^( 2) =t) (1/2)∫_0 ^( 1) (( t^( (i/2)) βˆ’t^((i+1)/2) )/(1βˆ’t))dt = (1/2) { ψ (1 +((i+1)/2))βˆ’Οˆ (1+(i/2))} = (1/2) { (2/(1+i)) + ψ ( ((1+i)/2) )βˆ’(2/i)βˆ’Οˆ ((i/2))} = ((βˆ’i)/(βˆ’1+i)) + (1/2) {ψ (((1+i)/2))βˆ’Οˆ((i/2) )} = βˆ’(1/2) +(i/(2 )) +(1/2) {ψ(((1+i)/2))βˆ’Οˆ((i/2))} Ξ© = Im (Ξ¦ )= (1/2) +(1/2) Im(ψ((1/2) +(i/2)))βˆ’(1/2) Im((i/2)) = (1/(2 )) { 1 +(Ο€/2) tanh((Ο€/2)) βˆ’1βˆ’(Ο€/2) coth((Ο€/2))} =(Ο€/4) {((βˆ’1)/(sinh((Ο€/2)).cosh((Ο€/2))))}=((βˆ’Ο€)/(2 sinh (Ο€ ))) βœ“

$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\Omega\::=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{x}.{sin}\left({ln}\left({x}\right)\right)}{\mathrm{1}βˆ’{x}}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:{method}\:\mathrm{1} \\ $$$$\:\:\:\:\:\Omega=\:\mathrm{I}{m}\left[\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\:{x}^{\:{i}+\mathrm{1}} }{\mathrm{1}βˆ’{x}}\:{dx}=\Phi\right] \\ $$$$\:\:\:\:\:\:\:\:\:\Phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{x}^{\:{i}+\mathrm{1}} +{x}^{\:{i}+\mathrm{2}} }{\mathrm{1}βˆ’{x}^{\:\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{{x}^{\:\mathrm{2}} ={t}} {=}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{t}^{\:\frac{{i}}{\mathrm{2}}} βˆ’{t}^{\frac{{i}+\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}βˆ’{t}}{dt} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left\{\:\psi\:\left(\mathrm{1}\:+\frac{{i}+\mathrm{1}}{\mathrm{2}}\right)βˆ’\psi\:\left(\mathrm{1}+\frac{{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left\{\:\:\frac{\mathrm{2}}{\mathrm{1}+{i}}\:+\:\psi\:\left(\:\frac{\mathrm{1}+{i}}{\mathrm{2}}\:\right)βˆ’\frac{\mathrm{2}}{{i}}βˆ’\psi\:\left(\frac{{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:=\:\frac{βˆ’{i}}{βˆ’\mathrm{1}+{i}}\:\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:\left\{\psi\:\left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)βˆ’\psi\left(\frac{{i}}{\mathrm{2}}\:\right)\right\} \\ $$$$\:\:\:=\:βˆ’\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{{i}}{\mathrm{2}\:}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\left\{\psi\left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)βˆ’\psi\left(\frac{{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\Omega\:=\:\mathrm{I}{m}\:\left(\Phi\:\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{I}{m}\left(\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{{i}}{\mathrm{2}}\right)\right)βˆ’\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{I}{m}\left(\frac{{i}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}\:}\:\left\{\:\mathrm{1}\:+\frac{\pi}{\mathrm{2}}\:{tanh}\left(\frac{\pi}{\mathrm{2}}\right)\:βˆ’\mathrm{1}βˆ’\frac{\pi}{\mathrm{2}}\:{coth}\left(\frac{\pi}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:=\frac{\pi}{\mathrm{4}}\:\:\left\{\frac{βˆ’\mathrm{1}}{{sinh}\left(\frac{\pi}{\mathrm{2}}\right).{cosh}\left(\frac{\pi}{\mathrm{2}}\right)}\right\}=\frac{βˆ’\pi}{\mathrm{2}\:{sinh}\:\left(\pi\:\right)}\:\checkmark \\ $$$$ \\ $$

Question Number 154186    Answers: 1   Comments: 0

Question Number 154177    Answers: 1   Comments: 2

Question Number 154175    Answers: 2   Comments: 1

Question Number 154172    Answers: 0   Comments: 1

Question Number 154148    Answers: 1   Comments: 0

z^4 - ((50)/(2z^4 - 7)) = 14 β‡’ z = ?

$$\boldsymbol{{z}}^{\mathrm{4}} \:-\:\frac{\mathrm{50}}{\mathrm{2}\boldsymbol{{z}}^{\mathrm{4}} \:-\:\mathrm{7}}\:=\:\mathrm{14}\:\:\:\Rightarrow\:\:\:\boldsymbol{{z}}\:=\:? \\ $$

Question Number 154143    Answers: 2   Comments: 0

Question Number 154142    Answers: 2   Comments: 0

Question Number 154138    Answers: 1   Comments: 0

Question Number 154133    Answers: 2   Comments: 0

Given f(x)=((x+(√(1+x^2 ))))^(1/3) +((xβˆ’(√(1+x^2 ))))^(1/3) Find f^(βˆ’1) (x)=?

$$\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt[{\mathrm{3}}]{\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}+\sqrt[{\mathrm{3}}]{\mathrm{x}βˆ’\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }} \\ $$$$\mathrm{Find}\:\mathrm{f}^{βˆ’\mathrm{1}} \left(\mathrm{x}\right)=? \\ $$

Question Number 154131    Answers: 0   Comments: 1

A long distance runner runs 14km 30Β°north of east, 7km 60Β° north of west, 6km 30Β° south of west, and finally 4km south. Find his final distance and direction relative to the starting point?

$$ \\ $$A long distance runner runs 14km 30Β°north of east, 7km 60Β° north of west, 6km 30Β° south of west, and finally 4km south. Find his final distance and direction relative to the starting point?

Question Number 154200    Answers: 1   Comments: 0

g(((xβˆ’1)/(x+1)))=((7x+3)/(x+1)) and f(x^2 βˆ’2x+3)=3x^2 βˆ’6x+7 find (f+g)(x)=?

$${g}\left(\frac{{x}βˆ’\mathrm{1}}{{x}+\mathrm{1}}\right)=\frac{\mathrm{7}{x}+\mathrm{3}}{{x}+\mathrm{1}}\:\:{and}\:\:{f}\left({x}^{\mathrm{2}} βˆ’\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{3}{x}^{\mathrm{2}} βˆ’\mathrm{6}{x}+\mathrm{7} \\ $$$${find}\:\:\left({f}+{g}\right)\left({x}\right)=?\:\:\: \\ $$

Question Number 154116    Answers: 0   Comments: 0

Question Number 154104    Answers: 3   Comments: 0

∫ e^(√x) dx =?

$$\:\:\int\:{e}^{\sqrt{{x}}} \:{dx}\:=? \\ $$

Question Number 154103    Answers: 2   Comments: 2

si w est une racine cubique de 1 different de 1,alors: (1+wβˆ’w^2 )^7 =?

$${si}\:{w}\:{est}\:{une}\:{racine}\:{cubique}\:{de}\:\mathrm{1}\:{different}\:{de}\:\mathrm{1},{alors}: \\ $$$$\left(\mathrm{1}+{w}βˆ’{w}^{\mathrm{2}} \right)^{\mathrm{7}} =? \\ $$

Question Number 154102    Answers: 0   Comments: 0

Question Number 154100    Answers: 1   Comments: 0

Question Number 154099    Answers: 1   Comments: 0

{ ((x^2 +y(√(xy)) = 72)),((y^2 +x(√(xy)) = 36)) :}

$$\:\begin{cases}{{x}^{\mathrm{2}} +{y}\sqrt{{xy}}\:=\:\mathrm{72}}\\{{y}^{\mathrm{2}} +{x}\sqrt{{xy}}\:=\:\mathrm{36}}\end{cases} \\ $$

Question Number 154088    Answers: 1   Comments: 1

Question Number 154087    Answers: 0   Comments: 0

Question Number 154085    Answers: 1   Comments: 1

Question Number 154081    Answers: 1   Comments: 0

lim_(xβ†’βˆž) ((32x^5 βˆ’14x^4 +3))^(1/5) βˆ’((128x^7 +6x^6 βˆ’1))^(1/7) =?

$$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{5}}]{\mathrm{32}{x}^{\mathrm{5}} βˆ’\mathrm{14}{x}^{\mathrm{4}} +\mathrm{3}}βˆ’\sqrt[{\mathrm{7}}]{\mathrm{128}{x}^{\mathrm{7}} +\mathrm{6}{x}^{\mathrm{6}} βˆ’\mathrm{1}}\:=? \\ $$

Question Number 154080    Answers: 1   Comments: 0

Ξ© =∫_0 ^( (Ο€/2)) ln^2 (((1+sin t)/(1βˆ’sin t)))dt

$$\:\:\:\:\Omega\:=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \mathrm{ln}\:^{\mathrm{2}} \left(\frac{\mathrm{1}+\mathrm{sin}\:{t}}{\mathrm{1}βˆ’\mathrm{sin}\:{t}}\right){dt} \\ $$

  Pg 646      Pg 647      Pg 648      Pg 649      Pg 650      Pg 651      Pg 652      Pg 653      Pg 654      Pg 655   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com