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dx-e-2x-5e-x-




Question Number 86518 by john santu last updated on 29/Mar/20
∫  (dx/(e^(2x) −5e^x ))
$$\int\:\:\frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{2x}} −\mathrm{5e}^{\mathrm{x}} } \\ $$
Commented by john santu last updated on 29/Mar/20
⇒ ∫  (dx/(e^x  (e^x  −5))) = ∫ ((e^x  dx)/(e^(2x)  (e^x −5)))  let t = e^x  −5 ⇒ dt = e^x  dx  ∫ (dt/((t+5)^2  t)) = ∫ (A/(t+5)) dt + ∫ (B/((t+5)^2 )) dt  + ∫ (C/t) dt
$$\Rightarrow\:\int\:\:\frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{x}} \:\left(\mathrm{e}^{\mathrm{x}} \:−\mathrm{5}\right)}\:=\:\int\:\frac{\mathrm{e}^{\mathrm{x}} \:\mathrm{dx}}{\mathrm{e}^{\mathrm{2x}} \:\left(\mathrm{e}^{\mathrm{x}} −\mathrm{5}\right)} \\ $$$$\mathrm{let}\:\mathrm{t}\:=\:\mathrm{e}^{\mathrm{x}} \:−\mathrm{5}\:\Rightarrow\:\mathrm{dt}\:=\:\mathrm{e}^{\mathrm{x}} \:\mathrm{dx} \\ $$$$\int\:\frac{\mathrm{dt}}{\left(\mathrm{t}+\mathrm{5}\right)^{\mathrm{2}} \:\mathrm{t}}\:=\:\int\:\frac{\mathrm{A}}{\mathrm{t}+\mathrm{5}}\:\mathrm{dt}\:+\:\int\:\frac{\mathrm{B}}{\left(\mathrm{t}+\mathrm{5}\right)^{\mathrm{2}} }\:\mathrm{dt} \\ $$$$+\:\int\:\frac{\mathrm{C}}{\mathrm{t}}\:\mathrm{dt}\: \\ $$$$ \\ $$

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