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Question Number 102461 by bemath last updated on 09/Jul/20

$$\int \frac{dx}{\sqrt{5e^{2x}+4e^x+1}}=?$$

Answered by PRITHWISH SEN 2 last updated on 09/Jul/20

$$\int \frac{\mathrm{dx}}{{\mathrm{e}}^{\mathrm{x}}\sqrt{\frac{1}{{\mathrm{e}}^{2\mathrm{x}}}+\frac{4}{{\mathrm{e}}^{\mathrm{x}}}+5}}\mathrm{put}\frac{1}{{\mathrm{e}}^{\mathrm{x}}}=\mathrm{t}\Rightarrow \frac{\mathrm{dx}}{{\mathrm{e}}^{\mathrm{x}}}=-\mathrm{dt}$$$$=-\int \frac{\mathrm{dt}}{\sqrt{{\mathrm{t}}^2+4\mathrm{t}+5}}=-\int \frac{\mathrm{dt}}{\sqrt{{(\mathrm{t}+2)}^2+1}}$$$$=-\ln \mid (\mathrm{t}+2)+\sqrt{{\mathrm{t}}^2+4\mathrm{t}+5}\mid +\mathrm{C}$$$$=\mathbf{ln}\mid \frac{{\mathbf{e}}^{\mathbf{x}}}{2{\mathbf{e}}^{\mathbf{x}}+1+\sqrt{5{\mathbf{e}}^{2\mathbf{x}}+4{\mathbf{e}}^{\mathbf{x}}+1}}\mid +\mathbf{C}$$$$\mathbf{please}\mathbf{check}.$$

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