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Question Number 154192 Answers: 1 Comments: 0

Ω :=∫_0 ^( 1) (( x.sin(ln(x)))/(1−x))dx method 1 Ω= Im[∫_0 ^( 1) (( x^( i+1) )/(1−x)) dx=Φ] Φ = ∫_0 ^( 1) (( x^( i+1) +x^( i+2) )/(1−x^( 2) ))dx =^(x^( 2) =t) (1/2)∫_0 ^( 1) (( t^( (i/2)) −t^((i+1)/2) )/(1−t))dt = (1/2) { ψ (1 +((i+1)/2))−ψ (1+(i/2))} = (1/2) { (2/(1+i)) + ψ ( ((1+i)/2) )−(2/i)−ψ ((i/2))} = ((−i)/(−1+i)) + (1/2) {ψ (((1+i)/2))−ψ((i/2) )} = −(1/2) +(i/(2 )) +(1/2) {ψ(((1+i)/2))−ψ((i/2))} Ω = Im (Φ )= (1/2) +(1/2) Im(ψ((1/2) +(i/2)))−(1/2) Im((i/2)) = (1/(2 )) { 1 +(π/2) tanh((π/2)) −1−(π/2) coth((π/2))} =(π/4) {((−1)/(sinh((π/2)).cosh((π/2))))}=((−π)/(2 sinh (π ))) ✓

$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\Omega\::=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{x}.{sin}\left({ln}\left({x}\right)\right)}{\mathrm{1}−{x}}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:{method}\:\mathrm{1} \\ $$$$\:\:\:\:\:\Omega=\:\mathrm{I}{m}\left[\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\:{x}^{\:{i}+\mathrm{1}} }{\mathrm{1}−{x}}\:{dx}=\Phi\right] \\ $$$$\:\:\:\:\:\:\:\:\:\Phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{x}^{\:{i}+\mathrm{1}} +{x}^{\:{i}+\mathrm{2}} }{\mathrm{1}−{x}^{\:\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{{x}^{\:\mathrm{2}} ={t}} {=}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{t}^{\:\frac{{i}}{\mathrm{2}}} −{t}^{\frac{{i}+\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}−{t}}{dt} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left\{\:\psi\:\left(\mathrm{1}\:+\frac{{i}+\mathrm{1}}{\mathrm{2}}\right)−\psi\:\left(\mathrm{1}+\frac{{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left\{\:\:\frac{\mathrm{2}}{\mathrm{1}+{i}}\:+\:\psi\:\left(\:\frac{\mathrm{1}+{i}}{\mathrm{2}}\:\right)−\frac{\mathrm{2}}{{i}}−\psi\:\left(\frac{{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:=\:\frac{−{i}}{−\mathrm{1}+{i}}\:\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:\left\{\psi\:\left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)−\psi\left(\frac{{i}}{\mathrm{2}}\:\right)\right\} \\ $$$$\:\:\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{{i}}{\mathrm{2}\:}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\left\{\psi\left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)−\psi\left(\frac{{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\Omega\:=\:\mathrm{I}{m}\:\left(\Phi\:\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{I}{m}\left(\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{{i}}{\mathrm{2}}\right)\right)−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{I}{m}\left(\frac{{i}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}\:}\:\left\{\:\mathrm{1}\:+\frac{\pi}{\mathrm{2}}\:{tanh}\left(\frac{\pi}{\mathrm{2}}\right)\:−\mathrm{1}−\frac{\pi}{\mathrm{2}}\:{coth}\left(\frac{\pi}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:=\frac{\pi}{\mathrm{4}}\:\:\left\{\frac{−\mathrm{1}}{{sinh}\left(\frac{\pi}{\mathrm{2}}\right).{cosh}\left(\frac{\pi}{\mathrm{2}}\right)}\right\}=\frac{−\pi}{\mathrm{2}\:{sinh}\:\left(\pi\:\right)}\:\checkmark \\ $$$$ \\ $$

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 ⇒ 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) +((x−(√(1+x^2 ))))^(1/3) Find f^(−1) (x)=?

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

Question Number 154131 Answers: 0 Comments: 1

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

$$ \\ $$A long distance runner runs 14km 30°north of east, 7km 60° north of west, 6km 30° 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(((x−1)/(x+1)))=((7x+3)/(x+1)) and f(x^2 −2x+3)=3x^2 −6x+7 find (f+g)(x)=?

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

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+w−w^2 )^7 =?

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

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_(x→∞) ((32x^5 −14x^4 +3))^(1/5) −((128x^7 +6x^6 −1))^(1/7) =?

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

Question Number 154080 Answers: 1 Comments: 0

Ω =∫_0 ^( (π/2)) ln^2 (((1+sin t)/(1−sin t)))dt

$$\:\:\:\:\Omega\:=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \mathrm{ln}\:^{\mathrm{2}} \left(\frac{\mathrm{1}+\mathrm{sin}\:{t}}{\mathrm{1}−\mathrm{sin}\:{t}}\right){dt} \\ $$

Question Number 154078 Answers: 0 Comments: 1

Question Number 154068 Answers: 3 Comments: 3

Question Number 154065 Answers: 1 Comments: 0

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