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Question Number 99072 by Mr.D.N. last updated on 18/Jun/20

Answered by mr W last updated on 18/Jun/20

∫((x^2 +2x−1)/(2x^2 +3x+1))dx =(1/2)∫((2x^2 +4x−2)/(2x^2 +3x+1))dx =(1/2)∫((2x^2 +3x+1+x−3)/(2x^2 +3x+1))dx =(1/2)∫[1+(1/4)(((4x+3−15)/(2x^2 +3x+1)))]dx =(1/2)∫[1+(1/4)(((4x+3)/(2x^2 +3x+1)))−((15)/2)×(1/((2x+1)(2x+2)))]dx =(1/2){∫dx+(1/4)∫((d(2x^2 +3x+1))/(2x^2 +3x+1))−((15)/4)×∫((1/(2x+1))−(1/(2x+2)))d(2x)} =(1/2){x+(1/4)ln (2x^2 +3x+1)−((15)/4)ln ((2x+1)/(2(x+1)))}+C =(x/2)+(1/8)ln (2x+1)(x+1)−((15)/8)ln ((2x+1)/(x+1))+C =(x/2)−(7/4)ln (2x+1)+2 ln (x+1)+C

$$\int\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{2}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}+{x}−\mathrm{3}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{4}{x}+\mathrm{3}−\mathrm{15}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}}\right)\right]{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{4}{x}+\mathrm{3}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}}\right)−\frac{\mathrm{15}}{\mathrm{2}}×\frac{\mathrm{1}}{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{2}\right)}\right]{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\int{dx}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{d}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\right)}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}}−\frac{\mathrm{15}}{\mathrm{4}}×\int\left(\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{2}}\right){d}\left(\mathrm{2}{x}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{x}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\right)−\frac{\mathrm{15}}{\mathrm{4}}\mathrm{ln}\:\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}\right\}+{C} \\ $$$$=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\:\left(\mathrm{2}{x}+\mathrm{1}\right)\left({x}+\mathrm{1}\right)−\frac{\mathrm{15}}{\mathrm{8}}\mathrm{ln}\:\frac{\mathrm{2}{x}+\mathrm{1}}{{x}+\mathrm{1}}+{C} \\ $$$$=\frac{{x}}{\mathrm{2}}−\frac{\mathrm{7}}{\mathrm{4}}\mathrm{ln}\:\left(\mathrm{2}{x}+\mathrm{1}\right)+\mathrm{2}\:\mathrm{ln}\:\left({x}+\mathrm{1}\right)+{C} \\ $$

Commented by Mr.D.N. last updated on 18/Jun/20

awesome thank you mr. W keep rocking��������✍��

Commented by niroj last updated on 18/Jun/20

what grate deal.��

Answered by maths mind last updated on 18/Jun/20

cos(x)(y′′−2tg(x)y′−(a^2 +1)y=(1/(cos(x)))e^x )...E⇔ hello i think that sec=(1/(cos(x))) ....we not use it in french sorry my engilish is not good cos(x)y′′−2sin(x)y′−(a^2 +1)cos(x)y=e^x ...E let z=cos(x)y ⇒z′=cos(x)y′−sin(x)y z′′=−2sin(x)y′−cos(x)y+cos(x)y′′ z′′−a^2 z=e^x ,easy now try sir is mor helpfull for you than give all answer find Z than use y=(z/(cos(x))) i hop i help you verry nice day for all of you

$${cos}\left({x}\right)\left({y}''−\mathrm{2}{tg}\left({x}\right){y}'−\left({a}^{\mathrm{2}} +\mathrm{1}\right){y}=\frac{\mathrm{1}}{{cos}\left({x}\right)}{e}^{{x}} \right)...{E}\Leftrightarrow \\ $$$${hello}\:{i}\:{think}\:{that}\:\:{sec}=\frac{\mathrm{1}}{{cos}\left({x}\right)}\:\:....{we}\:{not}\:{use}\:{it}\:{in}\:{french}\: \\ $$$${sorry}\:{my}\:{engilish}\:{is}\:{not}\:{good}\: \\ $$$${cos}\left({x}\right){y}''−\mathrm{2}{sin}\left({x}\right){y}'−\left({a}^{\mathrm{2}} +\mathrm{1}\right){cos}\left({x}\right){y}={e}^{{x}} ...{E} \\ $$$${let}\:{z}={cos}\left({x}\right){y} \\ $$$$\Rightarrow{z}'={cos}\left({x}\right){y}'−{sin}\left({x}\right){y} \\ $$$${z}''=−\mathrm{2}{sin}\left({x}\right){y}'−{cos}\left({x}\right){y}+{cos}\left({x}\right){y}'' \\ $$$${z}''−{a}^{\mathrm{2}} {z}={e}^{{x}} \:\:\:\:,{easy}\:{now}\:{try}\:{sir}\:{is}\:{mor}\:{helpfull}\:{for}\:{you}\: \\ $$$${than}\:{give}\:{all}\:{answer} \\ $$$${find}\:{Z}\:{than}\:{use}\:{y}=\frac{{z}}{{cos}\left({x}\right)}\:\: \\ $$$${i}\:{hop}\:{i}\:{help}\:{you}\:{verry}\:{nice}\:{day}\:{for}\:{all}\:{of}\:{you} \\ $$$$ \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 18/Jun/20

let solve z^(′′) −a^2 z =e^x (he)→r^2 −a^2 =0 ⇒r =+^− a if a is real z_h =α e^(ax) +β e^(−ax) =αu_1 +βu_2 W(u_1 ,u_2 ) = determinant (((e^(ax) e^(−ax) )),((ae^(ax) −a e^(−ax) )))=−2a (we suppose a≠0) w_1 = determinant (((0 e^(−ax) )),((e^x −ae^(−ax) )))=−e^((1−a)x) W_2 = determinant (((e^(ax) 0)),((ae^x e^x )))=e^((a+1)x) v_1 =∫ (W_1 /W)dx =−∫ (e^((1−a)x) /(−2a)) dx =(1/(2a(1−a)))e^((1−a)x) v_2 =∫ (W_2 /W)dx =∫ (e^((a+1)x) /(−2a)) =−(1/(2a(a+1))) e^((a+1)x) Z_p =u_1 v_1 +u_2 v_2 =e^(ax) ×(1/(2a(1−a)))e^((1−a)x) −e^(−ax) ×(1/(2a(a+1)))e^((a+1)x) =(e^x /(2a(1−a))) −(e^x /(2a(1+a))) =(e^x /(2a))((1/(1−a))−(1/(1+a))) =(e^x /(2a))(((2a)/(1−a^2 ))) =(e^x /(1−a^2 )) the general solution is z =z_h +z_p =αe^(ax) +β e^(−ax) +(e^x /(1−a^2 )) (a^2 ≠1)

$$\mathrm{let}\:\mathrm{solve}\:\mathrm{z}^{''} −\mathrm{a}^{\mathrm{2}} \mathrm{z}\:=\mathrm{e}^{\mathrm{x}} \:\:\:\:\left(\mathrm{he}\right)\rightarrow\mathrm{r}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \:=\mathrm{0}\:\Rightarrow\mathrm{r}\:=\overset{−} {+}\mathrm{a}\:\:\mathrm{if}\:\mathrm{a}\:\mathrm{is}\:\mathrm{real}\: \\ $$$$\mathrm{z}_{\mathrm{h}} =\alpha\:\mathrm{e}^{\mathrm{ax}} \:+\beta\:\mathrm{e}^{−\mathrm{ax}} \:\:\:=\alpha\mathrm{u}_{\mathrm{1}} \:+\beta\mathrm{u}_{\mathrm{2}} \\ $$$$\mathrm{W}\left(\mathrm{u}_{\mathrm{1}} ,\mathrm{u}_{\mathrm{2}} \right)\:=\begin{vmatrix}{\mathrm{e}^{\mathrm{ax}} \:\:\:\:\:\:\:\:\:\:\mathrm{e}^{−\mathrm{ax}} }\\{\mathrm{ae}^{\mathrm{ax}} \:\:\:\:\:\:\:\:−\mathrm{a}\:\mathrm{e}^{−\mathrm{ax}} }\end{vmatrix}=−\mathrm{2a}\:\:\:\:\:\:\:\left(\mathrm{we}\:\mathrm{suppose}\:\mathrm{a}\neq\mathrm{0}\right) \\ $$$$\mathrm{w}_{\mathrm{1}} =\begin{vmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{e}^{−\mathrm{ax}} }\\{\mathrm{e}^{\mathrm{x}} \:\:\:\:\:\:\:\:−\mathrm{ae}^{−\mathrm{ax}} }\end{vmatrix}=−\mathrm{e}^{\left(\mathrm{1}−\mathrm{a}\right)\mathrm{x}} \\ $$$$\mathrm{W}_{\mathrm{2}} =\begin{vmatrix}{\mathrm{e}^{\mathrm{ax}} \:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{ae}^{\mathrm{x}} \:\:\:\:\:\:\:\:\:\mathrm{e}^{\mathrm{x}} }\end{vmatrix}=\mathrm{e}^{\left(\mathrm{a}+\mathrm{1}\right)\mathrm{x}} \\ $$$$\mathrm{v}_{\mathrm{1}} =\int\:\frac{\mathrm{W}_{\mathrm{1}} }{\mathrm{W}}\mathrm{dx}\:=−\int\:\:\frac{\mathrm{e}^{\left(\mathrm{1}−\mathrm{a}\right)\mathrm{x}} }{−\mathrm{2a}}\:\mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{2a}\left(\mathrm{1}−\mathrm{a}\right)}\mathrm{e}^{\left(\mathrm{1}−\mathrm{a}\right)\mathrm{x}} \\ $$$$\mathrm{v}_{\mathrm{2}} =\int\:\frac{\mathrm{W}_{\mathrm{2}} }{\mathrm{W}}\mathrm{dx}\:=\int\:\frac{\mathrm{e}^{\left(\mathrm{a}+\mathrm{1}\right)\mathrm{x}} }{−\mathrm{2a}}\:=−\frac{\mathrm{1}}{\mathrm{2a}\left(\mathrm{a}+\mathrm{1}\right)}\:\mathrm{e}^{\left(\mathrm{a}+\mathrm{1}\right)\mathrm{x}} \\ $$$$\mathrm{Z}_{\mathrm{p}} =\mathrm{u}_{\mathrm{1}} \mathrm{v}_{\mathrm{1}} \:+\mathrm{u}_{\mathrm{2}} \mathrm{v}_{\mathrm{2}} =\mathrm{e}^{\mathrm{ax}} ×\frac{\mathrm{1}}{\mathrm{2a}\left(\mathrm{1}−\mathrm{a}\right)}\mathrm{e}^{\left(\mathrm{1}−\mathrm{a}\right)\mathrm{x}} \:−\mathrm{e}^{−\mathrm{ax}} \:×\frac{\mathrm{1}}{\mathrm{2a}\left(\mathrm{a}+\mathrm{1}\right)}\mathrm{e}^{\left(\mathrm{a}+\mathrm{1}\right)\mathrm{x}} \\ $$$$=\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{2a}\left(\mathrm{1}−\mathrm{a}\right)}\:−\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{2a}\left(\mathrm{1}+\mathrm{a}\right)}\:=\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{2a}}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{a}}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{a}}\right)\:=\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{2a}}\left(\frac{\mathrm{2a}}{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }\right)\:=\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{1}−\mathrm{a}^{\mathrm{2}} } \\ $$$$\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{z}\:=\mathrm{z}_{\mathrm{h}} \:+\mathrm{z}_{\mathrm{p}} =\alpha\mathrm{e}^{\mathrm{ax}} \:+\beta\:\mathrm{e}^{−\mathrm{ax}} \:\:+\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }\:\:\:\left(\mathrm{a}^{\mathrm{2}} \:\neq\mathrm{1}\right) \\ $$

Commented by Mr.D.N. last updated on 18/Jun/20

thanks all of you sir for your best effort Mr.mathmax your nearest answering . Ans: y= sec x ( C_1 e^(ax) +C_2 e^(−ax) + (e^x /(1−a^2 )))//.

$$\:\mathrm{thanks}\:\mathrm{all}\:\mathrm{of}\:\mathrm{you}\:\mathrm{sir}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{your}}\:\boldsymbol{\mathrm{best}}\:\boldsymbol{\mathrm{effort}}\:\mathrm{Mr}.\mathrm{mathmax}\:\:\mathrm{your}\:\mathrm{nearest}\:\mathrm{answering}\:\:. \\ $$$$\:\:\mathrm{Ans}:\:\mathrm{y}=\:\mathrm{sec}\:\mathrm{x}\:\left(\:\mathrm{C}_{\mathrm{1}} \mathrm{e}^{\mathrm{ax}} +\mathrm{C}_{\mathrm{2}} \mathrm{e}^{−\mathrm{ax}} +\:\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }\right)//. \\ $$

Commented by mathmax by abdo last updated on 18/Jun/20

you are welcome sir.

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir}. \\ $$

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