How do I do this? (Integration)?

Question

Ok,

I'm starting to figure out the integration thing with regular numbers, etc. but how do you integrate natural log equations?

This is what I have:

∫[ln(2x)]^2/2x dx

I know that
∫x^-1 dx= ∫1/x dx = dx/x = ln x + C, x>0 and
∫x^-1 dx= ln |x| +C, x<0

But how do I do it in reverse order? Can anyone help me?

Answer 1

It depends on how you are going about integration.

The substitution method is fine and will get you there eventually but in many cases it is a very slow method. These three answers are fine.

In many integrals where substitution works there is a much quicker way for one type where the integral is of the form
S f(x)f '(x)dx where S is the integral sign.

fox example S cosx sin^3 x dx = sin^4x / 4 + c because d/dx(sin^4x) = 4sin^3xcosx

In your question d/dx(ln2x) = 2/2x so to work out your integral you just increase the power by one from 2 to 3 and adjust the multiplier to make the numbers correct

The answer will be [ln(2x)]^3 give or take a number

d/dx([ln(2x)^3] = 2. 3. [ln(2x)]^2/2x

The answer is {[ln(2x)]^3}/6 + c

This might look complicated written out like this but with practice it is a very quick and powerful method. It has the advantage of only involving differentiation which is much easier because there are set methods unlike integration which requires a measure of recognition

Answer 2

You need to use u-substitution for this one.
Let u = ln(2x).
Then du = 2/(2x)dx = 1/x dx
so 1/2 du = 1/(2x) dx

The integral becomes integral of 1/2 u^2 du, which you can integrate using the power rule.

Answer 3

Substitute u for ln(2x). Then du = dx/x, and you have
(1/2)∫u^2du =
(1/6)u^3 + C =
(1/6[ln(2n)]^3 + C

Answer 4

Let u = ln(2x)

du/dx= 2* 1/(2x) = 1/x
du = (1/x) dx

Your integral simplifies to

∫(u^2)/2 du
=(u^3)/6 + c
= {[ln(2x)]^3}/6 + c

Answer 5

I = ∫ [ln (2x)]² / 2x dx
Let u = 2x
du/dx = 2 and dx = du / 2
I = (1/2). ∫ [ ln u ]² / u du
Let w = ln u
dw = (1/u) du
I = (1/2) ∫ w² dw
I = (1/6) w³ + C
I = (1/6).(ln u) ³ + C
I = (1/6).(ln 2x)³ + C

Answer 6

This type problem is done by the process known as "integration by parts". It is a very powerful method, and almost half of integral calculus is working with this process. The basic equation is
S U dV = U V - S V dU. where "S" is an integral sign.