Question

Find the integral: $\int {\dfrac{{{e^x}(1 + x\log x)}}{x}dx}$.

Answer 1

Hint – Divide by x in the numerator and denominator and then assume $f(x) = \log x$. The integral will transform into $\int {{e^x}[f(x) + f'(x)]dx}$ form and the integration of this form is $\int {{e^x}[f(x) + f'(x)]dx} = {e^x}f(x) + c$.

Complete step-by-step answer: -
Let us assume the given integral as,
$I = \int {\dfrac{{{e^x}(1 + x\log x)}}{x}dx}$.
Now, dividing by x in numerator and denominator, we get-

$$I = \int {\dfrac{{{e^x}(1 + x\log x)}}{x}dx} \\\ I = \int {{e^x}\left( {\dfrac{1}{x} + \log x} \right)} dx \\\$$

Now, if we consider, $f(x) = \log x$, then $f'(x) = \dfrac{1}{x}$.
So, we can write the above integral I as-
$I = \int {{e^x}[f(x) + f'(x)]dx}$
Now, we know that $\int {{e^x}[f(x) + f'(x)]dx} = {e^x}f(x) + c$.
Therefore, the above integral value is-
$I = \int {{e^x}[f(x) + f'(x)]dx} = {e^x}f(x) + c$
Put the value $f(x) = \log x$ in the above integral, I we get-
$I = {e^x}\log x + c$.
Hence, the integration of $\int {\dfrac{{{e^x}(1 + x\log x)}}{x}dx} = {e^x}\log x + c$.

Note – Whenever this type of question appears then as mentioned in the solution, assume the integral as $I$, divide both numerator and denominator by x. After dividing, we can see that $f(x) = \log x$ and $f'(x) = \dfrac{1}{x}$. So, the above integral can be written in the form $I = \int {{e^x}[f(x) + f'(x)]dx}$ and we know the integration of such forms is $\int {{e^x}[f(x) + f'(x)]dx} = {e^x}f(x) + c$, and then again keeping the value of f(x) as log x in I , we get the integration value.