Impossible evaluation NEED HELP T_T

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Homework Statement


evaluate the integral of 25x^2lnxdx



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The Attempt at a Solution


First thing I did was set 25x^2 as D, then set lnx as I. Derivative of 25x^2 is 50x. The antiderivative of lnx is xlnx-x. Therefore it turned out to be something like...25x^2(xlnx-x)-integral of 50x(xlnx-x). Then i proceeded by finding the antiderivative of my second integral...but soon found out that its an unending process if this keeps going up. PLEASE HELP. ADVICE IS REALLY APPRECIATED. THANK YOU SO MUCH.
 
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You need to use substitution twice. Start with In (x) as D
 
rootX said:
You need to use substitution twice. Start with In (x) as D

alright Ill try that, thanks for the advice. And by the way do you perhaps know the antiderivative of e^(ln(4)*x)? Because on this one certain problem i had to find the antiderivative of 4^x which i found to be e^(ln(4)*x) but i can't seem to know how to find the antiderivative of e^(ln(4)*x). Is it perhaps e^(ln(4)*x) all over ln(4)x?

thank you for taking your time to help me out. Really appreciate it.
 
Isn't this like the third thread out of 5 top ones that ask the same question?
 
MathNoob123 said:
First thing I did was set 25x^2 as D, then set lnx as I. Derivative of 25x^2 is 50x. The antiderivative of lnx is xlnx-x. Therefore it turned out to be something like...25x^2(xlnx-x)-integral of 50x(xlnx-x). Then i proceeded by finding the antiderivative of my second integral...but soon found out that its an unending process if this keeps going up. PLEASE HELP. ADVICE IS REALLY APPRECIATED. THANK YOU SO MUCH.

There's a nice little trick here if you realize you have:

\int 25x^2\ln x dx=25x^2(x\ln x -x)-\int50x^2\ln x dx +\int 50x^2 dx

\implies \int 50x^2 \ln x dx+ \int 25x^2\ln x dx=3\int 25x^2\ln x dx=25x^2(x\ln x -x)+\int 50x^2 dx

\implies\int 25x^2\ln x dx=\frac{1}{3}\left(25x^2(x\ln x -x)+\int 50x^2 dx\right)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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