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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

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

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

$$cos\left(x\right)(y^{″}-2tg\left(x\right)y^{\prime }-(a^2+1)y=\frac{1}{cos\left(x\right)}{e}^x)...E\iff$$$$helloithinkthatsec=\frac{1}{cos\left(x\right)}....wenotuseitinfrench$$$$sorrymyengilishisnotgood$$$$cos\left(x\right)y^{″}-2sin\left(x\right)y^{\prime }-(a^2+1)cos\left(x\right)y=e^x...E$$$$letz=cos\left(x\right)y$$$$\Rightarrow z^{\prime }=cos\left(x\right)y^{\prime }-sin\left(x\right)y$$$$z^{″}=-2sin\left(x\right)y^{\prime }-cos\left(x\right)y+cos\left(x\right)y^{″}$$$$z^{″}-a^{2}z=e^x,easynowtrysirismorhelpfullforyou$$$$thangiveallanswer$$$$findZthanusey=\frac{z}{cos\left(x\right)}$$$$ihopihelpyouverrynicedayforallofyou$$$$$$$$$$

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

$$\mathrm{let}\mathrm{solve}{\mathrm{z}}^{{}^{″}}-{\mathrm{a}}^2\mathrm{z}={\mathrm{e}}^{\mathrm{x}}\left(\mathrm{he}\right)\to {\mathrm{r}}^2-{\mathrm{a}}^2=0\Rightarrow \mathrm{r}=\stackrel{-}{+}\mathrm{a}\mathrm{if}\mathrm{a}\mathrm{is}\mathrm{real}$$$${\mathrm{z}}_{\mathrm{h}}=\alpha {\mathrm{e}}^{\mathrm{ax}}+\beta {\mathrm{e}}^{-\mathrm{ax}}=\alpha {\mathrm{u}}_1+\beta {\mathrm{u}}_2$$$$\mathrm{W}({\mathrm{u}}_1,{\mathrm{u}}_2)=\left|\begin{array}{c}{\mathrm{e}}^{\mathrm{ax}}{\mathrm{e}}^{-\mathrm{ax}}\\ {\mathrm{ae}}^{\mathrm{ax}}-\mathrm{a}{\mathrm{e}}^{-\mathrm{ax}}\end{array}\right|=-2\mathrm{a}(\mathrm{we}\mathrm{suppose}\mathrm{a}\ne 0)$$$${\mathrm{w}}_1=\left|\begin{array}{c}0{\mathrm{e}}^{-\mathrm{ax}}\\ {\mathrm{e}}^{\mathrm{x}}-{\mathrm{ae}}^{-\mathrm{ax}}\end{array}\right|=-{\mathrm{e}}^{(1-\mathrm{a})\mathrm{x}}$$$${\mathrm{W}}_2=\left|\begin{array}{c}{\mathrm{e}}^{\mathrm{ax}}0\\ {\mathrm{ae}}^{\mathrm{x}}{\mathrm{e}}^{\mathrm{x}}\end{array}\right|={\mathrm{e}}^{(\mathrm{a}+1)\mathrm{x}}$$$${\mathrm{v}}_1=\int \frac{{\mathrm{W}}_1}{\mathrm{W}}\mathrm{dx}=-\int \frac{{\mathrm{e}}^{(1-\mathrm{a})\mathrm{x}}}{-2\mathrm{a}}\mathrm{dx}=\frac{1}{2\mathrm{a}(1-\mathrm{a})}{\mathrm{e}}^{(1-\mathrm{a})\mathrm{x}}$$$${\mathrm{v}}_2=\int \frac{{\mathrm{W}}_2}{\mathrm{W}}\mathrm{dx}=\int \frac{{\mathrm{e}}^{(\mathrm{a}+1)\mathrm{x}}}{-2\mathrm{a}}=-\frac{1}{2\mathrm{a}(\mathrm{a}+1)}{\mathrm{e}}^{(\mathrm{a}+1)\mathrm{x}}$$$${\mathrm{Z}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2={\mathrm{e}}^{\mathrm{ax}}\times \frac{1}{2\mathrm{a}(1-\mathrm{a})}{\mathrm{e}}^{(1-\mathrm{a})\mathrm{x}}-{\mathrm{e}}^{-\mathrm{ax}}\times \frac{1}{2\mathrm{a}(\mathrm{a}+1)}{\mathrm{e}}^{(\mathrm{a}+1)\mathrm{x}}$$$$=\frac{{\mathrm{e}}^{\mathrm{x}}}{2\mathrm{a}(1-\mathrm{a})}-\frac{{\mathrm{e}}^{\mathrm{x}}}{2\mathrm{a}(1+\mathrm{a})}=\frac{{\mathrm{e}}^{\mathrm{x}}}{2\mathrm{a}}(\frac{1}{1-\mathrm{a}}-\frac{1}{1+\mathrm{a}})=\frac{{\mathrm{e}}^{\mathrm{x}}}{2\mathrm{a}}\left(\frac{2\mathrm{a}}{1-{\mathrm{a}}^2}\right)=\frac{{\mathrm{e}}^{\mathrm{x}}}{1-{\mathrm{a}}^2}$$$$\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{is}\mathrm{z}={\mathrm{z}}_{\mathrm{h}}+{\mathrm{z}}_{\mathrm{p}}=\alpha {\mathrm{e}}^{\mathrm{ax}}+\beta {\mathrm{e}}^{-\mathrm{ax}}+\frac{{\mathrm{e}}^{\mathrm{x}}}{1-{\mathrm{a}}^2}({\mathrm{a}}^2\ne 1)$$

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

$$\mathrm{thanks}\mathrm{all}\mathrm{of}\mathrm{you}\mathrm{sir}\mathbf{for}\mathbf{your}\mathbf{best}\mathbf{effort}\mathrm{Mr}.\mathrm{mathmax}\mathrm{your}\mathrm{nearest}\mathrm{answering}.$$$$\mathrm{Ans}:\mathrm{y}=\sec \mathrm{x}({\mathrm{C}}_1{\mathrm{e}}^{\mathrm{ax}}+{\mathrm{C}}_2{\mathrm{e}}^{-\mathrm{ax}}+\frac{{\mathrm{e}}^{\mathrm{x}}}{1-{\mathrm{a}}^2})//.$$

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

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

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