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Question-201172

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Question Number 201172 by Calculusboy last updated on 01/Dec/23
Answered by Sutrisno last updated on 01/Dec/23
=∫((2e^(2x) −e^x )/( (√(3(e^(2x) −2e^x −(1/3))))))dx =(1/( (√3)))∫((2e^(2x) −e^x )/( (√((e^x −1)^2 −(4/3)))))dx misal : e^x −1=(2/( (√3)))secθ→tanθ=((√(3e^(2x) −6e^x −1))/2) dx=(2/( (√3))).((secθ.tanθ)/e^x )dθ=(2/( (√3)))(((secθ.tanθ)/((2/( (√3)))secθ+1)))dθ =(1/( (√3)))∫((2((2/( (√3)))secθ+1)^2 −((2/( (√3)))secθ+1))/( (√(((2/( (√3)))secθ)^2 −(4/3))))).(2/( (√3)))(((secθ.tanθ)/((2/( (√3)))secθ+1)))dθ =(1/( (√3)))∫((2((2/( (√3)))secθ+1)−1)/( (2/( (√3))).tanθ)).(2/( (√3))).secθtanθdθ =(1/( (√3)))∫(4/( (√3)))sec^2 θ+secθdθ =(1/( (√3)))((4/( (√3)))tanθ+ln∣secθ+tanθ∣) =(1/( (√3)))((4/( (√3)))((√(3e^(2x) −6e^x −1))/2)+ln∣(((√3)(e^x −1))/2)+((√(3e^(2x) −6e^x −1))/2)∣)+c
$$=\int\frac{\mathrm{2}{e}^{\mathrm{2}{x}} −{e}^{{x}} }{\:\sqrt{\mathrm{3}\left({e}^{\mathrm{2}{x}} −\mathrm{2}{e}^{{x}} −\frac{\mathrm{1}}{\mathrm{3}}\right)}}{dx} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int\frac{\mathrm{2}{e}^{\mathrm{2}{x}} −{e}^{{x}} }{\:\sqrt{\left({e}^{{x}} −\mathrm{1}\right)^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{3}}}}{dx} \\ $$$${misal}\::\:{e}^{{x}} −\mathrm{1}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{sec}\theta\rightarrow{tan}\theta=\frac{\sqrt{\mathrm{3}{e}^{\mathrm{2}{x}} −\mathrm{6}{e}^{{x}} −\mathrm{1}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{dx}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}.\frac{{sec}\theta.{tan}\theta}{{e}^{{x}} }{d}\theta=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left(\frac{{sec}\theta.{tan}\theta}{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{sec}\theta+\mathrm{1}}\right){d}\theta \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int\frac{\mathrm{2}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{sec}\theta+\mathrm{1}\right)^{\mathrm{2}} −\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{sec}\theta+\mathrm{1}\right)}{\:\sqrt{\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{sec}\theta\right)^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{3}}}}.\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left(\frac{{sec}\theta.{tan}\theta}{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{sec}\theta+\mathrm{1}}\right){d}\theta \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int\frac{\mathrm{2}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{sec}\theta+\mathrm{1}\right)−\mathrm{1}}{\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}.{tan}\theta}.\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}.{sec}\theta{tan}\theta{d}\theta \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}{sec}^{\mathrm{2}} \theta+{sec}\theta{d}\theta \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}{tan}\theta+{ln}\mid{sec}\theta+{tan}\theta\mid\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\frac{\sqrt{\mathrm{3}{e}^{\mathrm{2}{x}} −\mathrm{6}{e}^{{x}} −\mathrm{1}}}{\mathrm{2}}+{ln}\mid\frac{\sqrt{\mathrm{3}}\left({e}^{{x}} −\mathrm{1}\right)}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}{e}^{\mathrm{2}{x}} −\mathrm{6}{e}^{{x}} −\mathrm{1}}}{\mathrm{2}}\mid\right)+{c} \\ $$$$ \\ $$$$ \\ $$
Commented by Calculusboy last updated on 04/Dec/23
thanks sir
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

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