How do I find the integral
#int(x*ln(x))dx# ?

Question

How do I find the integral
 #int(x*ln(x))dx# ?

Ezra Adams · Accepted Answer

Integration by parts will be used.

Recall the formula for the IBP, which is

#int u dv = uv - int v du#

Let #u = ln x#, and #dv = x dx#. We have chosen these values because we know that the derivative of #ln x# is equal to #1/x#, meaning that instead of integrating something complex (a natural logarithm) we now will end up integrating something pretty easy. (a polynomial)

Thus, #du = 1/x dx#, and #v = x^2 / 2#.

Entering the IBP formula provides us with:

#int x ln x dx = (x^2 ln x)/2 - int x^2 / (2x) dx#

An #x# will cancel off from the new integrand:

#int x ln x dx = (x^2 ln x)/2 - int x / 2 dx#

With the power rule, the solution can now be found with ease. Additionally, remember the integration constant:

#int x ln x dx = (x^2 ln x)/2 - x^2 / 4 + C#

William Abdalla · Answer

undefined To find the integral of $ \int x \ln(x) \, dx $, you can use integration by parts. Let $ u = \ln(x) $ and $ dv = x \, dx $. Then, $ du = \frac{1}{x} \, dx $ and $ v = \frac{1}{2}x^2 $.

Now apply the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

Substitute the values:

$$ \int x \ln(x) \, dx = \frac{1}{2}x^2\ln(x) - \int \frac{1}{2}x^2 \cdot \frac{1}{x} \, dx $$

$$ = \frac{1}{2}x^2\ln(x) - \frac{1}{2} \int x \, dx $$

$$ = \frac{1}{2}x^2\ln(x) - \frac{1}{4}x^2 + C $$

Where $ C $ is the constant of integration.

How do I find the integral #int(x*ln(x))dx# ?