What do you get when you integrate ln2x?

Answer 1

Remeber the natural log rules:
ln(a*b) = ln(a) + ln(b)
ln(a/b) = ln(a) - ln(b)
ln(a^b) = b*ln(a)

Break that ln(2x) down into ln(2) + ln(x).
int[ln(2) + ln(x)]dx = int[ln(2)*dx] + int[ln(x)*dx]

Since ln(2) is a constant, it'll just integrate into x*ln(2).

You would need to use integration by parts on int[ln(x)*dx].

In case you don't remember or haven't learned it yet, integration by parts go like this by product rule:
d(uv) = u*dv + v*du
int[d(uv)] = int[u*dv + v*du] = int[u*dv] + int[v*du]
--> uv = int[u*dv] + int[v*du]
--> int[u*dv] = uv - int[v*du]

Let u = ln(x) and dv = dx.
du = (1/x)dx
v = int(dv) = int(dx) = x
int[ln(x)dx] = x*ln(x) - int[x*(1/x)dx] = x*ln(x) - int(dx)
= x*ln(x) - x + C

int[ln(2x)dx] = int[ln(2) + ln(x)]dx = int[ln(2)*dx] + int[ln(x)*dx]
= x*ln(2) + x*ln(x) - x + C
=x[ln(2) + ln(x)] - x + C
=x[ln(2x)] - x + C
=ln[(2x)^x] - x + C

Answer 2

The answer is
x ( ln2x -1/2)

Answer 3

I will show each and every step as you requested. 2·∫ x·ln(2x) dx u = ln(2x) du = [1/x] dx dv = x dx v = ½·x² ∫ u dv = uv - ∫ v du 2·∫ x·ln(2x) dx = 2·[ ½·x²·ln(2x) - ∫ ½·x²·[1/x] dx 2·∫ x·ln(2x) dx = 2·[ ½·x²·ln(2x) - ½·∫ x dx] 2·∫ x·ln(2x) dx = 2·[ ½·x²·ln(2x) - ¼·x²] + C 2·∫ x·ln(2x) dx = x²·ln(2x) - ½·x² + C

Answer 4

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Answer 5

Let u = ln(2x), du = 2dx/x
dv = dx, v = x
∫ln(2x)dx = xln(2x) - ∫2xdx/x = xln(2x) - 2x + C

Answer 6

integration(ln(2x) dx) = (x*ln(2x)) - x + constant