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Indefinite integral with two parts

find_the_fun

Active member
Feb 1, 2012
166
I'm trying to integrate \(\displaystyle \int e^{4\ln{x}}x^2 dx\)
I can't see using u-substition, \(\displaystyle x^2\) isn't the derivative of \(\displaystyle e^{4\ln{x}}\) nor vice-versa.

I tried integrating by parts and that didn't work. I used \(\displaystyle u=e^{4\ln{x}}\) and \(\displaystyle dv=x^2 dx\)

I know I can't rewrite \(\displaystyle e^{4\ln{x}}\) as \(\displaystyle e^4e^\ln{x}\)
 
Last edited:

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Re: indefinite integral with two parts

I'm trying to integrate \(\displaystyle \int e^{4\ln{x}}x^2 dx\)
I can't see using u-substition, \(\displaystyle x^2\) isn't the derivative of \(\displaystyle e^{4\ln{x}}\) nor vice-versa.

I tried integrating by parts and that didn't work. I used \(\displaystyle u=e^{4\ln{x}}\) and \(\displaystyle dv=x^2 dx\)
Note that $e^{4\ln x} = e^{\ln(x^4)} = x^4$.

Can you take things from here?
 

find_the_fun

Active member
Feb 1, 2012
166
Re: indefinite integral with two parts

Note that $e^{4\ln x} = e^{\ln(x^4)} = x^4$.

Can you take things from here?
I guess the lesson learned from this is to simplify the expression algebraically before attempting integration techniques.