How do I evaluate the integral #int7x^2 ln(x) dx#?

1 Answer
KillerBunny
Jan 26, 2015

I'd use integration by parts: since #(fg)'=f'g+fg'#, by integrating this relation one has that #fg=\int f'g + \int fg'#.

When you have to integrate the product of two functions, integration by parts is often a good strategy: if one of the two factors is easily integrable/derivable, you can try and see if the integral becomes easier. In your case, I'd use the fact that
#(1) \int f'g=fg-\int fg'#

So, we have the following:
#f(x)= 7/3 x^3,\ \ \ f'(x)= 7x^2#
#g(x)=\ln(x),\ \ \ g'(x)=1/x#.

And using the equation labeled as #(1)#, you have that

#\int 7x^2 \ln(x) = 7/3 x^3 \ln(x) - \int 7/3 x^3 1/x#

The andantage is evident: the integral of #7/3 x^3 1/x=7/3 x^2# is obviously #7/9 x^3#, and so our solution is

#\int 7x^2 \ln(x) = 7/3 x^3 \ln(x) - 7/9 x^3+c#

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