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Question Number 197407 Answers: 2 Comments: 0

lim_(x→∞) ((√(x^2 +1))/(x+1))=?

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\mathrm{1}}=? \\ $$

Question Number 197396 Answers: 1 Comments: 1

Question Number 197393 Answers: 1 Comments: 0

Question Number 197389 Answers: 0 Comments: 0

please check my answer (x−2y+5)dx+(2x−y+4)dy=0 X=x+a & Y=y+b (X−a−2Y+2b+5)dX+(2X−2a−Y+b+4)dY=0 -a+2b=-5 -2a+b=-4 a=1,b=-2 (X−2Y)dX+(2X−Y)dY=0 Y=XV⇒(dY/dX)=V+X(dV/dx) (X−2XV)+(2X−XV)(V+X(dV/dX))=0 (2X−XV)(V+X(dV/dX))=-X+2XV (V+X(dV/dX))=((-1+2V)/(2−V)) X(dV/dX)=((-1+2V)/(2−V))−V X(dV/dX)=((-1+2V−2V+V^( 2) )/(2−V)) X(dV/dX)=((V^( 2) −1)/(2−V)) ∫((2−V)/(V^( 2) −1))dV=∫(dX/X) ∫(1/(2(V−1)))+(3/(2(V+1)))dV=ln x+C (1/2)ln(V−1)+(3/2)ln(V+1) (1/2)ln(((y−2)/(x+1))−1)+(3/2)ln(((y−2)/(x+1))+1)=ln x+C

$$ \\ $$$${please}\:{check}\:{my}\:{answer} \\ $$$$\:\left({x}−\mathrm{2}{y}+\mathrm{5}\right){dx}+\left(\mathrm{2}{x}−{y}+\mathrm{4}\right){dy}=\mathrm{0} \\ $$$$\:{X}={x}+{a}\:\&\:{Y}={y}+{b} \\ $$$$ \\ $$$$\:\left({X}−{a}−\mathrm{2}{Y}+\mathrm{2}{b}+\mathrm{5}\right){dX}+\left(\mathrm{2}{X}−\mathrm{2}{a}−{Y}+{b}+\mathrm{4}\right){dY}=\mathrm{0} \\ $$$$\:-{a}+\mathrm{2}{b}=-\mathrm{5} \\ $$$$\:-\mathrm{2}{a}+{b}=-\mathrm{4} \\ $$$$\:{a}=\mathrm{1},{b}=-\mathrm{2} \\ $$$$\:\left({X}−\mathrm{2}{Y}\right){dX}+\left(\mathrm{2}{X}−{Y}\right){dY}=\mathrm{0} \\ $$$$\:{Y}={XV}\Rightarrow\frac{{dY}}{{dX}}={V}+{X}\frac{{dV}}{{dx}} \\ $$$$\:\left({X}−\mathrm{2}{XV}\right)+\left(\mathrm{2}{X}−{XV}\right)\left({V}+{X}\frac{{dV}}{{dX}}\right)=\mathrm{0} \\ $$$$\:\left(\mathrm{2}{X}−{XV}\right)\left({V}+{X}\frac{{dV}}{{dX}}\right)=-{X}+\mathrm{2}{XV} \\ $$$$\:\left({V}+{X}\frac{{dV}}{{dX}}\right)=\frac{-\mathrm{1}+\mathrm{2}{V}}{\mathrm{2}−{V}} \\ $$$$\:{X}\frac{{dV}}{{dX}}=\frac{-\mathrm{1}+\mathrm{2}{V}}{\mathrm{2}−{V}}−{V} \\ $$$$\:{X}\frac{{dV}}{{dX}}=\frac{-\mathrm{1}+\mathrm{2}{V}−\mathrm{2}{V}+{V}^{\:\mathrm{2}} }{\mathrm{2}−{V}} \\ $$$$\:{X}\frac{{dV}}{{dX}}=\frac{{V}^{\:\mathrm{2}} −\mathrm{1}}{\mathrm{2}−{V}} \\ $$$$\:\int\frac{\mathrm{2}−{V}}{{V}^{\:\mathrm{2}} −\mathrm{1}}{dV}=\int\frac{{dX}}{{X}} \\ $$$$\:\int\frac{\mathrm{1}}{\mathrm{2}\left({V}−\mathrm{1}\right)}+\frac{\mathrm{3}}{\mathrm{2}\left({V}+\mathrm{1}\right)}{dV}={ln}\:{x}+{C} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({V}−\mathrm{1}\right)+\frac{\mathrm{3}}{\mathrm{2}}{ln}\left({V}+\mathrm{1}\right) \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{y}−\mathrm{2}}{{x}+\mathrm{1}}−\mathrm{1}\right)+\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\frac{{y}−\mathrm{2}}{{x}+\mathrm{1}}+\mathrm{1}\right)={ln}\:{x}+{C} \\ $$$$ \\ $$$$ \\ $$

Question Number 197388 Answers: 4 Comments: 0

Question Number 197461 Answers: 1 Comments: 0

Prove that: •∫^( x) _( 0) ((lnt)/(t^2 −1))dt=∫^( (π/2)) _( 0) arctan(xtanθ)dθ • ∫^( x) _( (1/x)) ((lnt)/(t^2 −1))arctant dt=(π/8)∫^( π) _( 0) arctan((1/2)(x−(1/x))sint)dt

$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\bullet\underset{\:\mathrm{0}} {\int}^{\:\mathrm{x}} \frac{\mathrm{lnt}}{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}\mathrm{dt}=\underset{\:\mathrm{0}} {\int}^{\:\frac{\pi}{\mathrm{2}}} \mathrm{arctan}\left(\mathrm{xtan}\theta\right)\mathrm{d}\theta \\ $$$$\bullet\:\:\underset{\:\frac{\mathrm{1}}{\mathrm{x}}} {\int}^{\:\mathrm{x}} \frac{\mathrm{lnt}}{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}\mathrm{arctant}\:\mathrm{dt}=\frac{\pi}{\mathrm{8}}\underset{\:\mathrm{0}} {\int}^{\:\pi} \mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{sint}\right)\mathrm{dt} \\ $$

Question Number 197374 Answers: 2 Comments: 0