Help with simple integration problem

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Solve y'= Ln(12x)



Homework Equations


Integral of ln(x)= xlnx-x dx


The Attempt at a Solution


12x*ln(12x)-12x dx or even xlnx-x*1/12
But I am pretty sure these are wrong.

My calculator gives me xlnx+(ln(12)-1)x wth? Help!
 
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do you know how to do integration by parts?
in order to solve this problem you need to use it.
 
CaityAnn said:
Solve y'= Ln(12x)



Homework Equations


Integral of ln(x)= xlnx-x dx


The Attempt at a Solution


12x*ln(12x)-12x dx or even xlnx-x*1/12
But I am pretty sure these are wrong.
Yup, they are incorrect.

To do this, you should note why the integral of ln(x) is equal to xlnx-x dx. Namely, we use integration by parts, writing the integrand as 1.ln(x). So, for your question, we have \int \ln(12x) dx=\int 1\cdot\ln(12x) dx. Now, can you see a suitable choice for "u" and "dv" to enable you to perform integration by parts on this?

My calculator gives me xlnx+(ln(12)-1)x wth?
Your calculator is correct. [When you have the answer, check to see if yours is the same as this]
 
Try substituting 12x = u and work from there.

EDIT: Damn my slowness. And also cristo's method is better, mine is rather lazy.
 
Last edited:
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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