Question

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

Answer 1

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. $\int {dx} = x + C$