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Question-183382

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Question Number 183382 by mathlove last updated on 25/Dec/22
Commented by CElcedricjunior last updated on 25/Dec/22
∫((x+1)/(2x^2 +6x+9))dx=(1/4)∫((4x+6)/(2x^2 +6x+9))dx−(1/4)∫((−2)/(2x^2 +6x+9))dx =(1/4)ln∣2x^2 +6x+9∣−(1/4)∫(dx/((x+(3/2))^2 +(9/4))) t=((2x+3)/3) =(1/4)ln∣2x^2 +6x+9∣−(1/6)arctan(((2x+3)/3))+c
$$\int\frac{\boldsymbol{{x}}+\mathrm{1}}{\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{6}\boldsymbol{{x}}+\mathrm{9}}\boldsymbol{{dx}}=\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{4}\boldsymbol{{x}}+\mathrm{6}}{\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{6}\boldsymbol{{x}}+\mathrm{9}}\boldsymbol{{dx}}−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{−\mathrm{2}}{\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{6}\boldsymbol{{x}}+\mathrm{9}}\boldsymbol{{dx}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{ln}}\mid\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{6}\boldsymbol{{x}}+\mathrm{9}\mid−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dx}}{\left(\boldsymbol{{x}}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{9}}{\mathrm{4}}}\:\boldsymbol{{t}}=\frac{\mathrm{2}\boldsymbol{{x}}+\mathrm{3}}{\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{\mathrm{ln}}\mid\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{6}\boldsymbol{{x}}+\mathrm{9}\mid−\frac{\mathrm{1}}{\mathrm{6}}\boldsymbol{{arctan}}\left(\frac{\mathrm{2}\boldsymbol{{x}}+\mathrm{3}}{\mathrm{3}}\right)+\boldsymbol{{c}} \\ $$
Commented by Acem last updated on 25/Dec/22
Salut, vous insistez pour mettre toute votre solutions en commentaires!
$$\:{Salut},\:{vous}\:{insistez}\:{pour}\:{mettre}\:{toute}\:{votre} \\ $$$$\:{solutions}\:{en}\:{commentaires}! \\ $$
Commented by mr W last updated on 26/Dec/22
To CElcedricjunior sir: last warning! When you insist on disobeying the rules of the forum and ignore the kind advises of other people, then this forum is not a suitable place for you, in other words, you are not welcome here, and I′ll ask the administrator to ban you from the forum! Image you are waiting for the bus at a bus stop in a queue, then a guy comes and pushes himself before all other people in the queue. Your current behavior is exactly the same as this guy. So please leave here or obey the rules here! if you have special reason for your doing, please let us know and speak it out (in French when you like). Just ignoring other people without saying any thing is not polite.
$${To}\:\boldsymbol{{CElcedricjunior}}\:{sir}: \\ $$$${last}\:{warning}! \\ $$$${When}\:{you}\:{insist}\:{on}\:{disobeying}\:{the}\: \\ $$$${rules}\:{of}\:{the}\:{forum}\:{and}\:{ignore}\:{the}\: \\ $$$${kind}\:{advises}\:{of}\:{other}\:{people},\:{then}\:{this}\: \\ $$$${forum}\:{is}\:{not}\:{a}\:{suitable}\:{place}\:{for}\:{you}, \\ $$$${in}\:{other}\:{words},\:{you}\:{are}\:{not}\:{welcome} \\ $$$${here},\:{and}\:{I}'{ll}\:{ask}\:{the}\:{administrator}\: \\ $$$${to}\:{ban}\:{you}\:{from}\:{the}\:{forum}! \\ $$$${Image}\:{you}\:{are}\:{waiting}\:{for}\:{the}\:{bus} \\ $$$${at}\:{a}\:{bus}\:{stop}\:{in}\:{a}\:{queue},\:{then}\:{a}\:{guy}\: \\ $$$${comes}\:{and}\:{pushes}\:{himself}\:{before}\:{all} \\ $$$${other}\:{people}\:{in}\:{the}\:{queue}.\:{Your} \\ $$$${current}\:{behavior}\:{is}\:{exactly}\:{the}\:{same} \\ $$$${as}\:{this}\:{guy}.\:{So}\:{please}\:{leave}\:{here}\:{or} \\ $$$${obey}\:{the}\:{rules}\:{here}! \\ $$$${if}\:{you}\:{have}\:{special}\:{reason}\:{for}\:{your} \\ $$$${doing},\:{please}\:{let}\:{us}\:{know}\:{and}\:{speak} \\ $$$${it}\:{out}\:\left({in}\:{French}\:{when}\:{you}\:{like}\right).\:{Just} \\ $$$${ignoring}\:{other}\:{people}\:{without}\:{saying} \\ $$$${any}\:{thing}\:{is}\:{not}\:{polite}. \\ $$
Commented by CElcedricjunior last updated on 26/Dec/22
i′m very sor
$${i}'{m}\:{very}\:{sor} \\ $$
Answered by Acem last updated on 25/Dec/22
I=∫ ((x+1)/(2x^2 +6x+9)) dx= (1/2)∫ ((x+1)/((x+(3/2))^2 +(9/4))) dx t= x+ (3/2) ... i I= (1/4) ∫ ((2t)/(t^2 +(9/4))) dt − (1/6) ∫ ((d((2/3) t))/(((2/3) t)^2 + 1)) I= (1/4) ln (x^2 +3x+(9/2)) − (1/6) arctan ((2/3) x +1) + c
$${I}=\int\:\frac{{x}+\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{9}}\:{dx}=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{x}+\mathrm{1}}{\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{9}}{\mathrm{4}}}\:{dx} \\ $$$$\:{t}=\:{x}+\:\frac{\mathrm{3}}{\mathrm{2}}\:…\:{i} \\ $$$$\:{I}=\:\frac{\mathrm{1}}{\mathrm{4}}\:\int\:\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\frac{\mathrm{9}}{\mathrm{4}}}\:{dt}\:−\:\frac{\mathrm{1}}{\mathrm{6}}\:\int\:\frac{{d}\left(\frac{\mathrm{2}}{\mathrm{3}}\:{t}\right)}{\left(\frac{\mathrm{2}}{\mathrm{3}}\:{t}\right)^{\mathrm{2}} +\:\mathrm{1}} \\ $$$$\:{I}=\:\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\frac{\mathrm{9}}{\mathrm{2}}\right)\:−\:\frac{\mathrm{1}}{\mathrm{6}}\:\mathrm{arctan}\:\left(\frac{\mathrm{2}}{\mathrm{3}}\:{x}\:+\mathrm{1}\right)\:\:+\:{c} \\ $$$$ \\ $$
Answered by Tony6400 last updated on 25/Dec/22
I=∫(x/(2x^2 +6x+9))dx+∫(1/(2x^2 +6x+9))dx=A+B Now A=∫(x/(2x^2 +6x+9))dx=∫(x/(2(x^2 +3x+(9/2))))dx=(1/2)∫(x/(x^2 +3x+((3/2))^2 +(9/2)−((3/2))^2 ))dx ⇒A=(1/2)∫(x/((x+(3/2))^2 +((3/2))^2 ))dx.Let x+(3/2)=(3/2)tanθ∴x=(3/2)(tanθ−1) ⇒(dx/dθ)=(3/2)sec^2 θ∴dx=(3/2)sec^2 θdθ ⇒A=(1/2)∫(((3/2)(tanθ−1))/(((3/2)tanθ)^2 +(9/4))).(3/2)sec^2 θdθ ∴A=(1/2)∫(((9/4)sec^2 θ(tanθ−1)dθ)/((9/4)(tan^2 θ+1)))=(1/2)∫(tanθ−1)dθ=(1/2)[−ln(cosθ)−θ].From x+(3/2)=(3/2)tanθ⇒tanθ=((2x+3)/3)=1+(2/3)x∴cosθ=(3/( (√((2x+3)^2 +9)))) And θ=arctan(1+(2/3)x)∴A=−(1/2)[ln((3/( (√((2x+3)^2 +9))))+arctan((2/3)x+1)]+C_1 Moreover B=∫(1/(2x^2 +6x+9))dx=(1/2)∫(1/((x+(3/2))^2 +((3/2))^2 ))dx=(1/2).(2/3)arctan(1+(2/3)x)+C_2 ⇒B=((arctan(1+(2/3)x))/3)+C_2 ⇒I=A+B=∫((x+1)/(2x^2 +6x+9))dx=−(1/2)[ln((3/( (√((2x+3)^2 +9)))))+arctan((2/3)x+1)]+((arctan(1+(2/3)x))/3)+K,where K=C_1 +C_2 I=−(1/2)ln((3/( (√((2x+3)^2 +9)))))−((arctan((2/3)x+1))/2)+((arctan(1+(2/3)x))/3)+K=−(1/2)ln((3/( (√((2x+3)^2 +9)))))−((arctan((2/3)x+1))/6)
$${I}=\int\frac{{x}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{9}}{dx}+\int\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{9}}{dx}={A}+{B} \\ $$$${Now}\:{A}=\int\frac{{x}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{9}}{dx}=\int\frac{{x}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\frac{\mathrm{9}}{\mathrm{2}}\right)}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{3}{x}+\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{9}}{\mathrm{2}}−\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx} \\ $$$$\Rightarrow{A}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}}{\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx}.{Let}\:{x}+\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}{tan}\theta\therefore{x}=\frac{\mathrm{3}}{\mathrm{2}}\left({tan}\theta−\mathrm{1}\right) \\ $$$$\Rightarrow\frac{{dx}}{{d}\theta}=\frac{\mathrm{3}}{\mathrm{2}}{sec}^{\mathrm{2}} \theta\therefore{dx}=\frac{\mathrm{3}}{\mathrm{2}}{sec}^{\mathrm{2}} \theta{d}\theta \\ $$$$\Rightarrow{A}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\frac{\mathrm{3}}{\mathrm{2}}\left({tan}\theta−\mathrm{1}\right)}{\left(\frac{\mathrm{3}}{\mathrm{2}}{tan}\theta\right)^{\mathrm{2}} +\frac{\mathrm{9}}{\mathrm{4}}}.\frac{\mathrm{3}}{\mathrm{2}}{sec}^{\mathrm{2}} \theta{d}\theta \\ $$$$\therefore{A}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\frac{\mathrm{9}}{\mathrm{4}}{sec}^{\mathrm{2}} \theta\left({tan}\theta−\mathrm{1}\right){d}\theta}{\frac{\mathrm{9}}{\mathrm{4}}\left({tan}^{\mathrm{2}} \theta+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\int\left({tan}\theta−\mathrm{1}\right){d}\theta=\frac{\mathrm{1}}{\mathrm{2}}\left[−{ln}\left({cos}\theta\right)−\theta\right].{From}\:{x}+\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}{tan}\theta\Rightarrow{tan}\theta=\frac{\mathrm{2}{x}+\mathrm{3}}{\mathrm{3}}=\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}{x}\therefore{cos}\theta=\frac{\mathrm{3}}{\:\sqrt{\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{9}}} \\ $$$${And}\:\theta={arctan}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}{x}\right)\therefore{A}=−\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left(\frac{\mathrm{3}}{\:\sqrt{\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{9}}}+{arctan}\left(\frac{\mathrm{2}}{\mathrm{3}}{x}+\mathrm{1}\right)\right]+{C}_{\mathrm{1}} \right. \\ $$$${Moreover}\:{B}=\int\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{9}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{2}}{\mathrm{3}}{arctan}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}{x}\right)+{C}_{\mathrm{2}} \\ $$$$\Rightarrow{B}=\frac{{arctan}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}{x}\right)}{\mathrm{3}}+{C}_{\mathrm{2}} \\ $$$$\Rightarrow{I}={A}+{B}=\int\frac{{x}+\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{9}}{dx}=−\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left(\frac{\mathrm{3}}{\:\sqrt{\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{9}}}\right)+{arctan}\left(\frac{\mathrm{2}}{\mathrm{3}}{x}+\mathrm{1}\right)\right]+\frac{{arctan}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}{x}\right)}{\mathrm{3}}+{K},{where}\:{K}={C}_{\mathrm{1}} +{C}_{\mathrm{2}} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{3}}{\:\sqrt{\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{9}}}\right)−\frac{{arctan}\left(\frac{\mathrm{2}}{\mathrm{3}}{x}+\mathrm{1}\right)}{\mathrm{2}}+\frac{{arctan}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}{x}\right)}{\mathrm{3}}+{K}=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{3}}{\:\sqrt{\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{9}}}\right)−\frac{{arctan}\left(\frac{\mathrm{2}}{\mathrm{3}}{x}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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