∫ xx(1 + log x)dx is equal to

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  1. xx + C
  2. x2x + C
  3. xx log x + C
  4. 1/2(1 + log x)2 + C 

Answer (Detailed Solution Below)

Option 1 : xx + C
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Detailed Solution

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Calculation

Let xˣ = t. Then,

⇒ d(xˣ) = dt

⇒ d(elog xx) = dt

⇒ d(ex log x) = dt

⇒ ex log x (log x + 1) dx = dt

⇒ xˣ (1 + log x) dx = dt

Therefore, I = xˣ (1 + log x) dx

⇒ I = dt = t + C = xˣ + C

Hence option 1 is correct

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