What is the integral of #xln^2(x+3)#?
1 Answer
I got
#(x^2 - 9)/2ln^2(x+3) - 1/2(x - 9)(x+3)ln(x+3) + 1/4(x+3)(x-21) + C#
DISCLAIMER: LONG ANSWER!
Whenever
For
#\mathbf(int udv = uv - intvdu)#
So the strategy here is to choose one term that is easier to differentiate, like
FIRST INTEGRATION BY PARTS
Let:
Now, you see how there's
SOME U SUBSTITUTION
Let
So, what we now have is:
#=> x^2/2ln^2(x+3) - int ((u-3)^2lnu)/udu#
#=> x^2/2ln^2(x+3) - int ((u^2 - 6u + 9)lnu)/udu#
#=> x^2/2ln^2(x+3) - int (u^2lnu - 6u lnu + 9lnu)/udu#
#=> x^2/2ln^2(x+3) - int u lnu - 6lnu + (9lnu)/udu#
That's not too bad now.
SECOND INTEGRATION BY PARTS
The first integral is done by three different integration by parts, but fairly straightforward ones.
For
For
#=> u^2/2 lnu - int u/2du = u^2/2 lnu - u^2/4#
THIRD INTEGRATION BY PARTS
For
#=> u lnu - int u/udu = u lnu - u#
FOURTH INTEGRATION BY PARTS
For
#=> 9int sds = 9/2s^2 = 9/2ln^2u#
PUTTING IT ALL TOGETHER
So, for our overall integral, we currently have:
#=> x^2/2ln^2(x+3) - [int u lnudu - int6lnudu + int(9lnu)/udu]#
#= x^2/2ln^2(x+3) - [(u^2/2 lnu - u^2/4) - 6(u lnu - u) + 9/2ln^2u]#
Make sure you catch those parentheses!
#= x^2/2ln^2(x+3) - [u^2/2 lnu - u^2/4 - 6u lnu + 6u + 9/2ln^2u]#
#= x^2/2ln^2(x+3) - u^2/2 lnu + u^2/4 + 6u lnu - 6u - 9/2ln^2u#
But since
#= x^2/2ln^2(x+3) - (x+3)^2/2 ln(x+3) + (x+3)^2/4 + 6(x+3) ln(x+3) - 6(x+3) - 9/2ln^2(x+3)#
Regroup terms together:
#= x^2/2ln^2(x+3) - 9/2ln^2(x+3) + 6(x+3) ln(x+3) - (x+3)^2/2 ln(x+3) + (x+3)^2/4 - 6(x+3)#
#= (x^2/2 - 9/2)ln^2(x+3) + (-(x+3)^2/2 + 6x + 18)ln(x+3) + (x+3)^2/4 - 6(x+3) + C#
Yeah, I think I'll stop here.
If you wanted to work to simplify this some more, you can get this eventually through some expansion and re-factoring:
#= color(blue)((x^2 - 9)/2ln^2(x+3) - 1/2(x - 9)(x+3)ln(x+3) + 1/4(x+3)(x-21) + C)#
Turns out that this was right!