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Question Number 166940 by cortano1 last updated on 03/Mar/22

Answered by som(math1967) last updated on 03/Mar/22

∫((e^x x+e^x )/(xe^x (1+xe^x )))dx  let  xe^x =z⇒(xe^x +e^x )dx=dz  ∫(dz/(z(z+1)))  =∫(dz/z) −∫(dz/(z+1))  =ln∣z∣ −ln∣z+1∣ +c   [z=xe^x ]

$$\int\frac{{e}^{{x}} {x}+{e}^{{x}} }{{xe}^{{x}} \left(\mathrm{1}+{xe}^{{x}} \right)}{dx} \\ $$$${let}\:\:{xe}^{{x}} ={z}\Rightarrow\left({xe}^{{x}} +{e}^{{x}} \right){dx}={dz} \\ $$$$\int\frac{{dz}}{{z}\left({z}+\mathrm{1}\right)} \\ $$$$=\int\frac{{dz}}{{z}}\:−\int\frac{{dz}}{{z}+\mathrm{1}} \\ $$$$={ln}\mid{z}\mid\:−{ln}\mid{z}+\mathrm{1}\mid\:+{c}\:\:\:\left[{z}={xe}^{{x}} \right] \\ $$

Commented by peter frank last updated on 03/Mar/22

thanks

$$\mathrm{thanks} \\ $$

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