Integrating ln(x)/4x: Steps and Tips for Solving

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Homework Statement



I am lost as to what to do here.

Homework Equations



Integral of (lnx)/(4x)

The Attempt at a Solution



let u = lnx
let du = (1/x)dx

(u)/(4x) dx...

But then howdo you make 4x disappear in the equation? Typically I did it by making du = something in the equation I want to take out, but how can you make 1/x = 4x?
 
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939 said:

Homework Statement



I am lost as to what to do here.

Homework Equations



Integral of (lnx)/(4x)

The Attempt at a Solution



let u = lnx
let du = (1/x)dx

(u)/(4x) dx...

But then howdo you make 4x disappear in the equation? Typically I did it by making du = something in the equation I want to take out, but how can you make 1/x = 4x?

Your integral is

$$\int dx~\frac{\ln x}{4x},$$
and you made the substitution u = ln x, so that du = dx/x, so you need to replace the dx in the integral with dx = x du. What happens to the 1/x in the integral then?
 
du = 1/x dx sooo

u/4 du...b/c...(1/4)*(ln(x)/x) dx
 
Brown Arrow said:
du = 1/x dx sooo

u/4 du...b/c...(1/4)*(ln(x)/x) dx

Sure, u/4 du. Integrate that.
 
Brown Arrow said:
du = 1/x dx sooo

u/4 du...b/c...(1/4)*(ln(x)/x) dx
And it would be a good idea to pull out that 1/4 right away so that you're working with this integral:
$$ \frac{1}{4} \int \frac{ln(x) dx}{x}$$
 
u = lnx, du = (1/x) dx

\frac{1}{4}∫u du

Then apply power rule...
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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