Question

Evaluate the value of $\int {{x^x}\left( {1 + \log x} \right)dx}$.

Answer 1

Hint:
Here, we are asked to find the value of $\int {{x^x}\left( {1 + \log x} \right)dx} $ .
Let $I = \int {{x^x}\left( {1 + \log x} \right)dx} $.
Then, let ${x^x} = t$ and differentiate on both sides and simplify the differentiation.
Thus, substitute the values and find the required answer.

Complete step by step solution:
Here, we are asked to find the value of $\int {{x^x}\left( {1 + \log x} \right)dx} $ .
Let $I = \int {{x^x}\left( {1 + \log x} \right)dx} $
To solve the given integral, let ${x^x} = t$ .
 $\therefore d\left( {{x^x}} \right) = dt$
Now, ${x^x}$ can also be written as ${e^{x\log x}}$ .
 $
  \therefore d\left( {{e^{x\log x}}} \right) = dt \\
  \therefore {e^{x\log x}}\left( {x \times \dfrac{1}{x} + \log x} \right) = dt \\
  \therefore {x^x}\left( {1 + \log x} \right)dx = dt \\
 $
Thus, $I = \int {dt} $
 $
  \therefore I = t + C \\
  \therefore I = {x^x} + C
 $

Thus, the value of the given integral is $\int {{x^x}\left( {1 + \log x} \right)dx} = {x^x} + C$.

Note:
The properties used in the question:
1) ${a^b} = {e^{b\log a}}$
2) $\dfrac{d}{{dx}}{e^{f\left( x \right)}} = {e^{f\left( x \right)}} \cdot f'\left( x \right)$
3) $\dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] = f\left( x \right)g'\left( x \right) + g\left( x \right)f'\left( x \right)$
4) $\int {dx} = x + C$