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Question Number 55316    Answers: 1   Comments: 1

Question Number 55312    Answers: 1   Comments: 0

Consider the system { ((x^2 + y^2 = z)),((2x + 2y + z = k)) :} The value of xy + zk for which the system has a unique solution is ...

$$\mathrm{Consider}\:\mathrm{the}\:\mathrm{system} \\ $$$$\begin{cases}{{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:=\:{z}}\\{\mathrm{2}{x}\:+\:\mathrm{2}{y}\:+\:{z}\:=\:{k}}\end{cases} \\ $$$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:{xy}\:+\:{zk}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{system} \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{unique}\:\mathrm{solution}\:\mathrm{is}\:... \\ $$

Question Number 55310    Answers: 1   Comments: 1

∫ 3x(√( 3x^3 + 7)) dx = . . . .

$$ \\ $$$$\int\:\mathrm{3}{x}\sqrt{\:\mathrm{3}{x}^{\mathrm{3}} +\:\mathrm{7}}\:\:{dx}\:=\:\:.\:.\:.\:. \\ $$

Question Number 55306    Answers: 1   Comments: 0

A mass of 6kg lies on an inclined plane which is smooth at angle θ to the horizontal where sin θ= (1/3).if it is connected to another mass 8kg by the same inelastic string passing over a smooth fixed pulley at the top of the plane.the partices are released from rest ,Find a) the acceleration of each mass b) the tension in the string c) the speed of the masses after 4seconds d) the distanced covered with this time in (c) above by the masses. 2017 CGCEB paper 3 Mechanics a/v

$${A}\:{mass}\:{of}\:\mathrm{6}{kg}\:{lies}\:{on}\:{an}\:{inclined}\:{plane} \\ $$$${which}\:{is}\:{smooth}\:{at}\:{angle}\:\theta\:{to}\:{the}\:{horizontal} \\ $$$${where}\:\:\mathrm{sin}\:\theta=\:\frac{\mathrm{1}}{\mathrm{3}}.{if}\:{it}\:{is}\:{connected}\:{to} \\ $$$${another}\:{mass}\:\mathrm{8}{kg}\:{by}\:{the}\:{same}\:{inelastic}\:{string} \\ $$$${passing}\:{over}\:{a}\:{smooth}\:{fixed}\:{pulley} \\ $$$${at}\:{the}\:{top}\:{of}\:{the}\:{plane}.{the}\:{partices}\:{are} \\ $$$${released}\:{from}\:{rest}\:,{Find}\: \\ $$$$\left.{a}\right)\:{the}\:{acceleration}\:{of}\:{each}\:{mass} \\ $$$$\left.{b}\right)\:{the}\:{tension}\:{in}\:{the}\:{string} \\ $$$$\left.{c}\right)\:{the}\:{speed}\:{of}\:{the}\:{masses}\:{after}\:\mathrm{4}{seconds} \\ $$$$\left.{d}\right)\:{the}\:{distanced}\:{covered}\:{with}\:{this}\:{time} \\ $$$${in}\:\left({c}\right)\:{above}\:{by}\:{the}\:{masses}. \\ $$$$ \\ $$$$\mathrm{2017}\:{CGCEB}\:{paper}\:\mathrm{3}\:{Mechanics}\:{a}/{v} \\ $$

Question Number 55301    Answers: 1   Comments: 2

find the fourth term in the expansion of (((√x)/y^2 )−(y/(√x)))^6

$${find}\:{the}\:{fourth}\:{term}\:{in}\:{the}\:{expansion} \\ $$$${of}\:\left(\frac{\sqrt{{x}}}{{y}^{\mathrm{2}} }−\frac{{y}}{\sqrt{{x}}}\right)^{\mathrm{6}} \\ $$

Question Number 55300    Answers: 1   Comments: 0

Question Number 55284    Answers: 0   Comments: 1

5^(x−2) −5^x +5^(x+1) =505

$$\mathrm{5}^{{x}−\mathrm{2}} −\mathrm{5}^{{x}} +\mathrm{5}^{{x}+\mathrm{1}} =\mathrm{505} \\ $$

Question Number 55282    Answers: 1   Comments: 1

calculatef(a)= ∫ (1+(a/x^2 ))arctan((a/x))dx 2) calculate ∫_1 ^(+∞) (1+(2/x^2 ))arctan((2/x))dx .

$${calculatef}\left({a}\right)=\:\:\int\:\:\:\left(\mathrm{1}+\frac{{a}}{{x}^{\mathrm{2}} }\right){arctan}\left(\frac{{a}}{{x}}\right){dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{1}} ^{+\infty} \left(\mathrm{1}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\right){arctan}\left(\frac{\mathrm{2}}{{x}}\right){dx}\:. \\ $$

Question Number 55281    Answers: 0   Comments: 0

find the value of Π_(n=1) ^∞ (1+(1/n^2 ))

$${find}\:{the}\:{value}\:{of}\:\prod_{{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right) \\ $$

Question Number 55280    Answers: 1   Comments: 1

fint f(t)=∫_0 ^1 ((ln(1+tx^2 ))/x^2 )dx .

$${fint}\:{f}\left({t}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{tx}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx}\:. \\ $$

Question Number 55279    Answers: 0   Comments: 0

find L( e^(−x) ln(1+x^2 )) with L mean laplace transform

$${find}\:{L}\left(\:{e}^{−{x}} {ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right)\:\:\:{with}\:{L}\:{mean}\:{laplace} \\ $$$${transform} \\ $$

Question Number 55278    Answers: 0   Comments: 0

calculate ∫_0 ^(+∞) (dx/((x+1)(x+2)....(x+n)))

$${calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)....\left({x}+{n}\right)} \\ $$

Question Number 55276    Answers: 0   Comments: 1

calculate Σ_(n=1) ^∞ (((−1)^n )/(16n^2 −1))

$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{16}{n}^{\mathrm{2}} −\mathrm{1}} \\ $$

Question Number 55274    Answers: 0   Comments: 4

let ϕ(a) =∫_1 ^(√3) arctan((a/x))dx 1) calculate ϕ(a) interms of a 2) calculate ϕ^′ (a) at form of integral. 3) determine ϕ^((n)) (a) at form of integral. 4) find the value of ∫_1 ^(√3) arctan((2/x))dx .

$${let}\:\varphi\left({a}\right)\:=\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:\:{arctan}\left(\frac{{a}}{{x}}\right){dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\varphi\left({a}\right)\:{interms}\:{of}\:{a} \\ $$$$\left.\mathrm{2}\right)\:\:{calculate}\:\varphi^{'} \left({a}\right)\:{at}\:{form}\:{of}\:{integral}. \\ $$$$\left.\mathrm{3}\right)\:{determine}\:\varphi^{\left({n}\right)} \left({a}\right)\:\:{at}\:{form}\:{of}\:{integral}. \\ $$$$\left.\mathrm{4}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \:{arctan}\left(\frac{\mathrm{2}}{{x}}\right){dx}\:. \\ $$

Question Number 55273    Answers: 0   Comments: 1

let f(x) =∫_x^2 ^(1+x) (dt/(1+t+t^2 )) 1) calculate f(x) interms of x 2) calculate lim_(x→0) f(x) and lim_(x→+∞) f(x)

$${let}\:\:{f}\left({x}\right)\:=\int_{{x}^{\mathrm{2}} } ^{\mathrm{1}+{x}} \:\:\frac{{dt}}{\mathrm{1}+{t}+{t}^{\mathrm{2}} } \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({x}\right)\:{interms}\:{of}\:{x} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)\:{and}\:{lim}_{{x}\rightarrow+\infty} \:\:{f}\left({x}\right) \\ $$

Question Number 55271    Answers: 0   Comments: 0

1) let f(x) =∫_0 ^(2π) ((cost)/(3 +sin(xt)))dt find a explicit form of f(x) 2) calculate g(x) =∫_0 ^(2π) ((tcos(xt)cost)/((3 +sin(xt))^2 ))dt 3) calculate ∫_0 ^(2π) ((cost)/(3+sint)) and ∫_0 ^(2π) ((t cos^2 t)/((3+sint)^2 ))dt

$$\left.\mathrm{1}\right)\:{let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{cost}}{\mathrm{3}\:+{sin}\left({xt}\right)}{dt} \\ $$$${find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{tcos}\left({xt}\right){cost}}{\left(\mathrm{3}\:+{sin}\left({xt}\right)\right)^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{cost}}{\mathrm{3}+{sint}}\:\:{and}\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{t}\:{cos}^{\mathrm{2}} {t}}{\left(\mathrm{3}+{sint}\right)^{\mathrm{2}} }{dt} \\ $$

Question Number 55270    Answers: 0   Comments: 1

let f(x)=x^n arctan(x^2 ) with n integr natural 1) calculate f^((n)) (x) and f^((n)) (0) 2)developp f at integr serie .

$${let}\:{f}\left({x}\right)={x}^{{n}} {arctan}\left({x}^{\mathrm{2}} \right)\:\:{with}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right){developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$$$ \\ $$

Question Number 55269    Answers: 0   Comments: 4

let f(x) =(2x+1)ln(1−x^2 ) 1) calculate f^((n)) (x) and f^((n)) (0) 2)developp f at integr serie. 3) calculate ∫_0 ^1 f(x)dx .

$${let}\:{f}\left({x}\right)\:=\left(\mathrm{2}{x}+\mathrm{1}\right){ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right){developp}\:{f}\:{at}\:{integr}\:{serie}. \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\:. \\ $$

Question Number 55268    Answers: 0   Comments: 1

calculate Σ_(n=0) ^∞ (((−1)^n )/(4n+3))

$${calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}+\mathrm{3}} \\ $$

Question Number 55308    Answers: 1   Comments: 0

Question Number 55288    Answers: 2   Comments: 0

Question Number 55258    Answers: 0   Comments: 0

3x+5y=?_

$$\mathrm{3}{x}+\mathrm{5}{y}=?_{} \\ $$

Question Number 55267    Answers: 0   Comments: 0

1) calculate f(x)=∫_0 ^(π/4) ln(1+xtanθ)dθ 2) find the values of integrals ∫_0 ^(π/4) ln(1+tanθ) and ∫_0 ^(π/4) ln(1+2tanθ)dθ . 1) we have f^′ (x)=∫_0 ^(π/4) ((tanθ)/(1+xtanθ)) dθ =∫_0 ^(π/4) (((sinθ)/(cosθ))/(1+x((sinθ)/(cosθ))))dθ =∫_0 ^(π/4) ((sinθ)/(cosθ +xsinθ)) dθ =_(tan((θ/2))=t) ∫_0 ^((√2)−1) (((2t)/(1+t^2 ))/(((1−t^2 )/(1+t^2 )) +((2xt)/(1+t^2 )))) ((2dt)/(1+t^2 )) =∫_0 ^((√2)−1) ((4t)/((1+t^2 )(1−t^2 +2xt)))dt =−∫_0 ^((√2)−1) ((4t)/((t^2 +1)(t^2 −2xt −1)))dt let decompose F(t) = ((4t)/((t^2 +1)(t^2 −2xt −1))) roots of t^2 −2xt −1 Δ^′ =x^2 +1 ⇒t_1 =x+(√(x^2 +1)) and t_2 =x−(√(x^2 +1)) F(t)=(a/(t−t_1 )) +(b/(t−t_2 )) +((ct +d)/(t^2 +1)) a =lim_(t→t_1 ) (t−t_1 )F(t)=((4t_1 )/((t_1 ^2 +1)(t_1 −t_2 ))) =α b =lim_(t→t_2 ) (t−t_2 )F(t) =((4t_2 )/((t_2 ^2 +1)(t_2 −t_1 ))) =β ⇒F(t)=(α/(t−t_1 )) +(β/(t−t_2 )) +((ct +d)/(t^2 +1)) F(0) =0=−(α/t_1 ) −(β/t_2 ) +d ⇒d =(α/t_1 ) +(β/t_2 ) F(1)=(2/(−2x)) =−(1/x)=(α/(1−t_1 )) +(β/(1−t_2 )) +((c+d)/2) ⇒(1/x) =(α/(t_1 −1)) +(β/(t_2 −1)) −(c/2) −(d/2) ⇒(c/2) =(α/(t_1 −1)) +(β/(t_2 −1)) −(d/2) −(1/x) ⇒c =((2α)/(t_1 −1)) +((2β)/(t_2 −1)) −d−(2/x) ∫ F(t)dt =αln∣t−t_1 ∣ +βln∣t−t_2 ∣ +(c/2)ln(t^2 +1) +d arctan(t) ⇒ ∫_0 ^((√2)−1) F(t)dt =[αln∣t−t_1 ∣+βln∣t−t_2 ∣ +(c/2)ln(t^2 +1)]_0 ^((√2)−1) =αln∣(√2)−1−t_1 ∣ +βln∣(√2)−1−t_2 ∣ +(c/2)ln(4−2(√2)) =αln∣(√2)−1−x−(√(1+x^2 )))+βln∣(√2)−1−x+(√(1+x^2 ))) +((ln(4−2(√2)))/2)c =f^′ (x) ⇒ f(x)=∫ αln∣(√2)−1−x−(√(1+x^2 ))∣)dx+β∫ ln∣(√2)−1+(√(1+x^2 ))∣dx +((cx)/2)ln(4−2(√2)) +C ....be continued...

$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left(\mathrm{1}+{xtan}\theta\right){d}\theta \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{values}\:{of}\:{integrals}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left(\mathrm{1}+{tan}\theta\right)\:\:{and}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+\mathrm{2}{tan}\theta\right){d}\theta\:. \\ $$$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{tan}\theta}{\mathrm{1}+{xtan}\theta}\:{d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{\frac{{sin}\theta}{{cos}\theta}}{\mathrm{1}+{x}\frac{{sin}\theta}{{cos}\theta}}{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{sin}\theta}{{cos}\theta\:+{xsin}\theta}\:{d}\theta\:\:=_{{tan}\left(\frac{\theta}{\mathrm{2}}\right)={t}} \:\:\:\:\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\:\frac{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:+\frac{\mathrm{2}{xt}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\:\:\frac{\mathrm{4}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}−{t}^{\mathrm{2}} \:+\mathrm{2}{xt}\right)}{dt}\:=−\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\:\frac{\mathrm{4}{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} −\mathrm{2}{xt}\:−\mathrm{1}\right)}{dt}\:{let}\:{decompose} \\ $$$${F}\left({t}\right)\:=\:\frac{\mathrm{4}{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} −\mathrm{2}{xt}\:−\mathrm{1}\right)}\:\:{roots}\:{of}\:\:{t}^{\mathrm{2}} −\mathrm{2}{xt}\:−\mathrm{1} \\ $$$$\Delta^{'} ={x}^{\mathrm{2}} +\mathrm{1}\:\Rightarrow{t}_{\mathrm{1}} ={x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:{and}\:{t}_{\mathrm{2}} ={x}−\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left({t}\right)=\frac{{a}}{{t}−{t}_{\mathrm{1}} }\:+\frac{{b}}{{t}−{t}_{\mathrm{2}} }\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${a}\:={lim}_{{t}\rightarrow{t}_{\mathrm{1}} } \left({t}−{t}_{\mathrm{1}} \right){F}\left({t}\right)=\frac{\mathrm{4}{t}_{\mathrm{1}} }{\left({t}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{1}\right)\left({t}_{\mathrm{1}} −{t}_{\mathrm{2}} \right)}\:=\alpha \\ $$$${b}\:={lim}_{{t}\rightarrow{t}_{\mathrm{2}} } \left({t}−{t}_{\mathrm{2}} \right){F}\left({t}\right)\:=\frac{\mathrm{4}{t}_{\mathrm{2}} }{\left({t}_{\mathrm{2}} ^{\mathrm{2}} \:+\mathrm{1}\right)\left({t}_{\mathrm{2}} −{t}_{\mathrm{1}} \right)}\:=\beta\:\Rightarrow{F}\left({t}\right)=\frac{\alpha}{{t}−{t}_{\mathrm{1}} }\:+\frac{\beta}{{t}−{t}_{\mathrm{2}} }\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{0}=−\frac{\alpha}{{t}_{\mathrm{1}} }\:−\frac{\beta}{{t}_{\mathrm{2}} }\:+{d}\:\:\Rightarrow{d}\:=\frac{\alpha}{{t}_{\mathrm{1}} }\:+\frac{\beta}{{t}_{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)=\frac{\mathrm{2}}{−\mathrm{2}{x}}\:=−\frac{\mathrm{1}}{{x}}=\frac{\alpha}{\mathrm{1}−{t}_{\mathrm{1}} }\:+\frac{\beta}{\mathrm{1}−{t}_{\mathrm{2}} }\:+\frac{{c}+{d}}{\mathrm{2}}\:\Rightarrow\frac{\mathrm{1}}{{x}}\:=\frac{\alpha}{{t}_{\mathrm{1}} −\mathrm{1}}\:+\frac{\beta}{{t}_{\mathrm{2}} −\mathrm{1}}\:−\frac{{c}}{\mathrm{2}}\:−\frac{{d}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{c}}{\mathrm{2}}\:=\frac{\alpha}{{t}_{\mathrm{1}} −\mathrm{1}}\:+\frac{\beta}{{t}_{\mathrm{2}} −\mathrm{1}}\:−\frac{{d}}{\mathrm{2}}\:−\frac{\mathrm{1}}{{x}}\:\Rightarrow{c}\:=\frac{\mathrm{2}\alpha}{{t}_{\mathrm{1}} −\mathrm{1}}\:+\frac{\mathrm{2}\beta}{{t}_{\mathrm{2}} −\mathrm{1}}\:−{d}−\frac{\mathrm{2}}{{x}} \\ $$$$\int\:{F}\left({t}\right){dt}\:=\alpha{ln}\mid{t}−{t}_{\mathrm{1}} \mid\:+\beta{ln}\mid{t}−{t}_{\mathrm{2}} \mid\:+\frac{{c}}{\mathrm{2}}{ln}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\:+{d}\:{arctan}\left({t}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} {F}\left({t}\right){dt}\:=\left[\alpha{ln}\mid{t}−{t}_{\mathrm{1}} \mid+\beta{ln}\mid{t}−{t}_{\mathrm{2}} \mid\:+\frac{{c}}{\mathrm{2}}{ln}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \\ $$$$=\alpha{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{t}_{\mathrm{1}} \mid\:+\beta{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{t}_{\mathrm{2}} \mid\:+\frac{{c}}{\mathrm{2}}{ln}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)\: \\ $$$$\left.=\left.\alpha{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)+\beta{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:+\frac{{ln}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{2}}{c}\:={f}^{'} \left({x}\right)\:\Rightarrow \\ $$$$\left.{f}\left({x}\right)=\int\:\alpha{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\mid\right){dx}+\beta\int\:{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\mid{dx} \\ $$$$+\frac{{cx}}{\mathrm{2}}{ln}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)\:+{C}\:....{be}\:{continued}... \\ $$

Question Number 55296    Answers: 1   Comments: 0

simplify the following expression if x<0 ∣4x−(√((3x−1)^2 ))∣

$${simplify}\:{the}\:{following}\:{expression}\:{if} \\ $$$${x}<\mathrm{0} \\ $$$$\mid\mathrm{4}{x}−\sqrt{\left(\mathrm{3}{x}−\mathrm{1}\right)^{\mathrm{2}} }\mid \\ $$

Question Number 55295    Answers: 0   Comments: 0

Question Number 55245    Answers: 1   Comments: 0

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