Question

# Find the value of $\int {\dfrac{{{e^x}}}{x}(x\log x + 1)dx}$.(a). $\dfrac{{{e^x}}}{x} + C$(b). $x{e^x}\log \left| x \right| + C$(c). ${e^x}\log \left| x \right| + C$(d). $x\left( {{e^x} + \log \left| x \right|} \right) + C$(e). $x{e^x} + \log \left| x \right| + C$

Hint: Separate the integrals into two terms separated by an addition. Simplify the first term using integration by parts and some terms will get cancelled to give the final answer.

The given integral has two terms separated by an addition. Let us make two integrals based on the rule of addition of integrals. Hence, we have:
$I = \int {\dfrac{{{e^x}}}{x}(x\log x + 1)dx}$
$I = \int {{e^x}\log xdx + \int {\dfrac{{{e^x}}}{x}dx} } ...........(1)$
Now, equation (1) has two parts, let's solve the first term to simplify the expression. Assign the first term to Iâ€™.
$I' = \int {{e^x}\log xdx}$
Let us use integration by parts to solve Iâ€™.
The formula for integration by parts is as follows:
$\int {udv = uv - \int {vdu} } ..........(2)$
We have, $u = \log x$ and $dv = {e^x}dx$. Hence, we find du and v as follows:
Find du by differentiating u as follows:
$du = \dfrac{1}{x}dx.........(3)$
Find v by integrating dv. We know that integration of ${e^x}$ is ${e^x}$ itself.
$\int {dv} = \int {{e^x}dx}$
$v = {e^x}............(4)$
Substituting equation (3) and equation (4) in equation (5), we have:
$\int {{e^x}\log xdx = \log \left| x \right|{e^x} - \int {\dfrac{{{e^x}}}{x}dx} } ..........(5)$
We now substitute equation (5) back in equation (1) to get:
$I = \log \left| x \right|{e^x} - \int {\dfrac{{{e^x}}}{x}dx} + \int {\dfrac{{{e^x}}}{x}dx}$
We can observe that the second and the third term cancel each other. Also, we need to add the constant of integration because the integral can differ by a constant. Hence, the final expression is as follows:
$I = {e^x}\log \left| x \right| + C$
Hence, the correct answer is ${e^x}\log \left| x \right| + C$.
Therefore, the correct answer is option (c).

Note: You must be careful when choosing u and v for integration by parts. A logarithmic function should be given a higher preference for u than the exponential function.