integration by parts integral of x(x+1)^5 dx
\[\int\limits x(x+1)^5\]
I think you might be better off by using a substitution, such as x+1=u.
@givanna618 , @Kainui is perfectly correct
Though even substituting u = x will work
But why bother using u then ?
You can always expand it. \((x+1)^{5} = x^{5} + 5x^{4} + 10x^{3} + 10x^{2} + 5x + 1\)
Yeah but that's really not necessary...
\[\large{\int x(x+1)^5 \ dx}\] \[\large{= \int (u-1)u^5 du}\] \[\large{= \int u^6 du - \int u^5 du}\] That is why @kainui suggested what he suggested
Using, u = x+1, ensures that you don't have to use binomial theorem :)
It gets easier that way
Great work @Kainui :D
Necessary? It's better than a brain cramp. It's absolutely the hardest way to do it, but it is a way to do it. If the alternative is to punt, why not do it a hard way?
Very wise words @tkhunny :)
Well @givanna618 now you have 2 different methods of solving the question thanks to @tkhunny and @Kainui . You can use whichever you like
And you can also use integration by parts
Though I would not prefer it ^^^
"by parts" was my third option.
:) By parts is a very general method. It can be used in many many integrals.
Sometimes I really wish there was an option of giving more than 1 medal.
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