is the integral of xln(x)dx=
x^2/4(4ln(x)-4-2ln(x)+3)+C?
To check--differentiate the integral to see if you get the original equation.
I believe you are correct.
You should simplify what you wrote. You have constants and terms in ln x that can be combined. If you do, you should end up with
x^2[(ln x)/2 - 1/4] + C
To find the integral of xln(x)dx, you can use integration by parts. The formula for integration by parts is as follows:
∫u dv = uv - ∫v du
Let's apply this formula:
Let u = ln(x)
du = (1/x) dx
Let dv = x dx
v = (1/2)x^2
Using the formula, we have:
∫xln(x)dx = ∫u dv
= uv - ∫v du
= (ln(x))(1/2)x^2 - ∫(1/2)x^2(1/x) dx
= (1/2)x^2ln(x) - (1/2)∫x dx
= (1/2)x^2ln(x) - (1/2)(1/2)x^2 + C
= (1/2)x^2ln(x) - (1/4)x^2 + C
Therefore, the correct answer should be (1/2)x^2ln(x) - (1/4)x^2 + C.