How Do I Integrate ln(4x) / 2x?

[metalmagik]

This integration might be very simple but I feel like I'm missing something here. I tried integration by parts but I ended up with a mess, x on top of x and the integral never came out without two x's as a product of each other.

$$\int$$$$\frac{ln(4x)}{2x}$$

Any help is appreciated, I'm trying to learn Calculus myself.

 

[maze]

Hint: what is d/dx(ln(x))?

 

[mmzaj]

let $$4x=y$$

=>

$$\int\frac{ln 4x}{2x}dx = \frac{1}{2}\int\frac{lny}{y} dy$$

integrate by parts :

$$lny = u , dv=\frac{dy}{y}$$

=>

$$du=\frac{dy}{y} , v=lny$$

=>

$$\int\frac{ln y}{y} dy = {(lny)}^2 - \int\frac{ln y}{y} dy$$

=>

$$\int\frac{ln y}{y} dy = \frac{1}{2} {(lny)}^2$$

=>


$$\int\frac{ln 4x}{2x}dx = \frac{1}{4} {(ln4x)}^2 +C$$

 

[Gib Z]

No, do NOT integrate by parts. Use maze's hint.

 

[mmzaj]

ok , maze's hint is smart , but i thought integrating by parts would be more informative as metalmagik said that he is learning !

 

[metalmagik]

Whoa, I didn't even know anyone replied to this yet.

Well mmzaj's strategy looks like it works, but d/dx of ln(x) is 1/x...

if I use u substitution for this problem, I get u = ln(4x), du = 1/x dx

So, I have to put a 2 outside the integral to balance.

2$$\int$$u du
2 $$\frac{u^2}{2}$$

Then the 2's cancel out and I'm left with u^2 or ln(4x)^2. But this is not what mmzaj got from integration by parts.

Unless my u-substitution is off...

Why is there a problem?

 

[exk]

f(x)=ln(u) f'(x)=u'/u

 

[maze]

The 2 should be in the denominator when you pull it out, and then multiplying by the 1/2 from u^2/2, that gets the factor of 1/4.