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Question Number 53311 by gunawan last updated on 20/Jan/19

$$\mathrm{If}\int \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}dx=Ax+B\log (9e^{2x}-4)+C$$$$\mathrm{then}$$$$A=...$$$$B=...$$$$C=...$$

Commented by maxmathsup by imad last updated on 20/Jan/19

$$letI=\int \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}dx\Rightarrow I{=}_{e^x=t}\int \frac{4t+6t^{-1}}{9t-4t^{-1}}\frac{dt}{t}=\int \frac{4t+6t^{-1}}{9t^2-4}dt$$$$=\int \frac{4t^2+6}{9t^3-4t}dtletdecomposeF\left(t\right)=\frac{4t^2+6}{t(9t^2-4)}=\frac{4t^2+6}{t(3t-2)(3t+2)}$$$$=\frac{a}{t}+\frac{b}{3t-2}+\frac{c}{3t+2}$$$$a={lim}_{t\to 0}tF\left(t\right)=-\frac{3}{2}$$$$b={lim}_{t\to \frac{2}{3}}(3t-2)F\left(t\right)=\frac{4.\frac{4}{9}+6}{\frac{2}{3}\left(4\right)}=\frac{\frac{8}{9}+3}{\frac{4}{3}}=\frac{35}{9}.\frac{3}{4}=\frac{35}{12}$$$$c={lim}_{t\to -\frac{2}{3}}(3t+2)F\left(t\right)=\frac{4.\frac{4}{9}+6}{(-\frac{2}{3})(-4)}=\frac{35}{12}\Rightarrow$$$$I=-\frac{3}{2}\int \frac{dt}{t}+\frac{35}{12}\int \frac{dt}{3t-2}+\frac{35}{12}\int \frac{dt}{3t+2}+c$$$$=-\frac{3}{2}ln\mid t\mid +\frac{35}{36}ln\mid 3t-2\mid +\frac{35}{36}ln\mid 3t+2\mid +c$$$$=-\frac{3}{2}ln\mid t\mid +\frac{35}{36}ln\mid 9t^2-4\mid +c$$$$=-\frac{3}{2}x+\frac{35}{36}ln(9e^{2x}-4)+c\Rightarrow a=-\frac{3}{2}andb=\frac{35}{36}$$$$c\to constantofintegration.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19

$$4e^x+6e^{-x}=P(9e^x-4e^{-x})+Q\times \frac{d}{dx}(9e^x-4e^{-x})$$$$=P(9e^x-4e^{-x})+Q(9e^x+4e^{-x})$$$$=e^x(9P+9Q)+e^{-x}(-4P+4Q)$$$$9P+9Q=4\times 4$$$$-4P+4Q=6\times 9$$$$36P+36Q=16$$$$-36P+36Q=54$$$$72Q=70[Q=\frac{35}{36}]4P=4\times \frac{35}{36}-6$$$$4P=\frac{140-216}{36}[P=\frac{-76}{4\times 36}=\frac{-19}{36}]$$$$\int \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}dx$$$$=\int \frac{P(9e^x-4e^{-x})+Q\times \frac{d}{dx}(9e^x-4e^{-x})}{9e^x-4e^{-x}}dx$$$$=P\int dx+Q\int \frac{d(9e^x-4e^{-x})}{9e^x-4e^{-x}}$$$$=Px+Qln(9e^x-4e^{-x})+c_1$$$$=Px+Q\left[ln\left(\frac{9e^{2x}-4}{e^x}\right)\right]+c_1$$$$=Px+Qln(9e^{2x}-4)-Qln\left(e^x\right)+c_1$$$$=Px+Qln(9e^{2x}-4)-Qx+c_1$$$$=(P-Q)x+Qln(9e^{2x}-4)+c_1$$$$=(\frac{-19}{36}-\frac{35}{36})x+\frac{35}{36}ln(9e^{2x}-4)+c_1$$$$=\left(\frac{-54}{36}\right)x+\frac{35}{36}ln(9e^{2x}-4)+c_1$$$$A\to \frac{-54}{36}B\to \frac{35}{36}C\to c_1$$$$\mathit{p}\mathit{l}\mathit{s}\mathit{c}\mathit{h}\mathit{e}\mathit{c}\mathit{k}\mathit{s}\mathit{t}\mathit{e}\mathit{p}\mathit{s}\mathit{m}\mathit{i}\mathit{s}\mathit{t}\mathit{a}\mathit{k}\mathit{e}\mathit{i}\mathit{f}\mathit{a}\mathit{n}\mathit{y}$$$$$$$$$$$$$$$$$$

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