Evaluate the following integrals: (x+1)e^ x (xe^ x )dx

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Question

Evaluate the following integrals:

$\int(\text{x}+1)\text{e}^{\text{x}}\log(\text{xe}^{\text{x}})\text{dx}$

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Answer

$\int(\text{x}+1)\text{e}^{\text{x}}.\log(\text{xe}^{\text{x}})\text{dx}$

Let $\text{x e}^{\text{x}}=\text{t}$

$\Rightarrow\big(\text{x.e}^{\text{x}}+1.\text{e}^{\text{x}}\big)\text{dx=dt}$

$\therefore\int(\text{x}+1)\text{e}^{\text{x}}.\log(\text{x e}^\text{x})\text{dx}=\int1.\log (\text{t})\text{dt}$

$=\log\text{t}\int1\text{dt}-\int\big\{\frac{\text{d}}{\text{dt}}(\log\text{t)}-\int1\text{dt}\Big\}\text{dt}$

$=\log(\text{t})\times\text{t}-\int\frac{1}{\text{t}}\times\text{t dt}$

$=\text{t}\log(\text{t})-\text{t}+\text{C} \dots(1)$

Substituting the value of t in eq (1)

$\Rightarrow\int(\text{x}+1)\text{e}^{\text{x}}.\log(\text{x e}^{\text{x}})\text{dx} = (\text{x e}^{\text{x}}).\log(\text{x e}^{\text{x}})-\text{x e}^{\text{x}}+\text{C}$

$=\text{x e}^{\text{x}}\Big\{\log(\text{x e}^{\text{x}})-1\Big\}+\text{C}$

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