What does that substitution do to the integral?What is the resulting integral?

ana111790
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Homework Statement



Find the integral of:

[a*ln(b/(b -cx)) - kx] dx

Where all a,b,c,k are constants and x is the variable.

Homework Equations



The Attempt at a Solution


Rewrote is:
a*INT(ln(b/(b-cx)) dx) - k*INT(x dx)

I don't know how to solve the first part, (the second integral I know is kx^2/2)

Thank you for your help.
 
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I don't know this for you is helpful or not ?

You can check it out form this site.:smile:

http://www.tutorvista.com/math/integral-of-log-x
 
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You started attack and you have to continue.

∫a*(ln(b/(b-cx)) dx) = a*∫ln(b/(b-cx) dx

You have to remember basic property of logs! You can do this if you can do ∫ ln(b - cx) dx .

And you can do that if you can do ∫ln x dx .

If y = ln x , what does x = ?

What substitution for x does that suggest?
 
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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