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Excellent
nice solution! btw from 1:04 we can use one of Frullani's formulas with f(x) = exp(-x) and get the result right away
I shall look into it.
It is done
Hi. Just checked. Integrate 0 to ♾️ ln(bx)/(√x)(√x+1)(√ax+1) equals 1/2(ln(b^2/a)*integral above without log factor in the numerator. So, integral is zero only when b^2=a
I already started working on it. You shall see some thing soon.
It is done. I just posted the video about the integral you asked.
Since the derivative of ln(1-x) is -1/(1-x), shouldnt you need a negative to cancel the negative from the derivative?
That is correct. ln(1 - x))^(-2+1)/(-2+1) and we got our negative
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Excellent
nice solution! btw from 1:04 we can use one of Frullani's formulas with f(x) = exp(-x) and get the result right away
I shall look into it.
It is done
Hi. Just checked. Integrate 0 to ♾️ ln(bx)/(√x)(√x+1)(√ax+1) equals 1/2(ln(b^2/a)*integral above without log factor in the numerator. So, integral is zero only when b^2=a
I already started working on it. You shall see some thing soon.
It is done. I just posted the video about the integral you asked.
Since the derivative of ln(1-x) is -1/(1-x), shouldnt you need a negative to cancel the negative from the derivative?
That is correct. ln(1 - x))^(-2+1)/(-2+1) and we got our negative