Question

What is the integral of `(ln(xe^x))/x`?

Answer 1

`\int` `ln(xe^x)/(x)dx = ln^2(x)/2+x+C`

Explanation:

We are given:

`\int` `ln(xe^x)/(x)dx`

Using `ln(ab) = ln(a) + ln(b)`:

`=\int` `(ln(x) + ln(e^x))/(x)dx`

Using `ln(a^b) = bln(a)`:

`=\int` `(ln(x) + xln(e))/(x)dx`

Using `ln(e) = 1`:

`=\int` `(ln(x) + x)/(x)dx`

Splitting the fraction (`x/x = 1`):

`=\int` `(ln(x)/x + 1)dx`

Separating the summed integrals:

`=\int` `ln(x)/xdx + \int dx`

The second integral is simply `x + C`, where `C` is an arbitrary constant. The first integral, we use `u`-substitution:

Let `u \equiv ln(x)`, hence `du = 1/x dx`

Using `u`-substitution:

`=\int udu + x + C`

Integrating (the arbitrary constant `C` can absorb the arbitrary constant of the first indefinite integral:

`=u^2/2 + x + C`

Substituting back in terms of `x`:

`= ln^2(x)/2+x+C`

Answer 2

`int\ ln(xe^x)/x\ dx=ln^2(x)/2+x+C`

Explanation:

We begin by using the following logarithm identity:

`ln(ab)=ln(a)+ln(b)`

Applying this to the integral, we get:

`int\ (ln(xe^x))/x\ dx=int\ ln(x)/x+ln(e^x)/x\ dx=`

`=int\ ln(x)/x+x/x\ dx=int\ ln(x)/x+1\ dx=int\ ln(x)/x\ dx+x`

To evaluate the remaining integral, we use integration by parts:

`int\ f(x)g'(x)\ dx=f(x)g(x)-int\ f'(x)g(x)\ dx`

I will let `f(x)=ln(x)` and `g'(x)=1/x`. We can then compute that:

`f'(x)=1/x` and `g(x)=ln(x)`

We can then apply the integration by parts formula to get:

`int\ ln(x)/x\ dx=ln(x)*ln(x)-int\ ln(x)/x\ dx`

Since we have the integral on both sides of the equals sign, we can solve it like an equation:

`2int\ ln(x)/x\ dx=ln^2(x)`

`int\ ln(x)/x\ dx=ln^2(x)/2+C`

Plugging back into the original expression, we get our final answer:

`int\ ln(xe^x)/x\ dx=ln^2(x)/2+x+C`