What's the indefinite integral of ln(x)?

Question

ln(x) = natural log of x

x is a variable

Answer 1

Indefinite integrals are integrals without any limits on either side.
They are simply refered to as integrals.

http://www.efunda.com/math/integrals/integrals.cfm

Here you will integrals for almost everything!

Trust me, they work.

All the best.

Peace out.

Answer 2

I presume you want to solve

Integral (ln(x) dx)

How we would solve this problem is to use integration by parts. It doesn't look like two parts now, but imagine ln(x) being multiplied by 1. To make myself absolutely clear, I'll change the equation to:

Integral (ln(x) [1] dx)

Now to use integration by parts.
Let u = ln(x), dv = [1] dx
du = (1/x)dx, v = x

Now, our integral becomes uv - Integral (v(du)), OR

xln(x) - Integral [x(1/x) dx]

Notice that x times 1/x simplifies nicely to 1.

xln(x) - Integral [1 dx]

The integral of 1 is trivial; it's just x. So now we have

xln(x) - x + C

[When finding an indefinite integral or general antiderivative, we always have to add a constant C]

Is it *really* the solution? We can check by actually solving for the derivative.
Let f(x) = xln(x) - x + C. Then, using the product rule on xln(x),
f'(x) = [(1)ln(x) + (x)(1/x)] - 1
f'(x) = [ln(x) + 1] - 1
f'(x) = ln(x) + 1 - 1
f'(x) = ln(x) <============== correct!

Answer 3

The integral has been evaluated above, but it requires integration by parts so just in case you aren't familiar with that I'll explain.

Let's start with the product rule for derivatives. Given two functions u and v, we can say

d(uv) = udv + vdu
d(uv) - vdu = udv
next, integrate both sides (I'll use S as an integrand sign)

uv - S vdu = Sudv

which is integration by parts. so in this case, you have
S lnx dx
u = lnx dv = dx
du = 1/x dx v = x

uv - S vdu = Sudv
xlnx - S x(1/x)dx = S lnx dx
xlnx - Sdx = S lnx dx
xlnx - x + C = S lnx dx

Answer 4

∫ ln x dx

Assume
u = ln x and dv = dx.

therefore,
v = x and du = 1/x dx

Use integrating by parts:
∫ u dv = uv - ∫ v du

Therefore,
∫ ln x dx = ln x (x) - ∫ (x) (1/x dx)

Simplifying,
∫ ln x dx = x ln x - ∫ dx

Therefore, the indefinite integral of ln x is:
∫ ln x dx = x ln x - x + C

^_^

Answer 5

1/x

Answer 6

Int logxdx=1/loge[x(logx-1)]+C