Integral - an alternative to expanding the denominator?

[LikeMath]

Integral -- an alternative to expanding the denominator?

The following integral can be easily solved by expanding the denominator but I am wondering if there is a another way to solve it

\int\frac{1}{(x+1)^7-x^7-1}dx

 

[Vorde]

You could try doing integration by substitution, and then repeating it several (probably 7) times. But I'm sure you would either end up with something really ugly or encounter a problem pretty early on that stops you.

 

[DonAntonio]

The following integral can be easily solved by expanding the denominator but I am wondering if there is a another way to solve it

\int\frac{1}{(x+1)^7-x^7-1}dx

Would you be so kind as to show us how this integral is "easily" solved expanding whatever? I think this is a rather

horrible integral, expanding or not, and more than finding a way to make it more or less normal I'd love to see what's your way to solve it.

Thanx

DonAntonio

 

[LikeMath]

Would you be so kind as to show us how this integral is "easily" solved expanding whatever? I think this is a rather

horrible integral, expanding or not, and more than finding a way to make it more or less normal I'd love to see what's your way to solve it.

Thanx

DonAntonio

By the binomial theorem we get
(x+1)^7-x^7-1=7(x^6+3x^5+5x^4+3x^2+x)
now if we factorize this term we also get
x(x+1)(x^2+x+1)^2
then partial fraction completes the solution.

 

[DonAntonio]

By the binomial theorem we get
(x+1)^7-x^7-1=7(x^6+3x^5+5x^4+3x^2+x)
now if we factorize this term we also get
x(x+1)(x^2+x+1)^2
then partial fraction completes the solution.

Well, yes...but for this you must first (1) know how to factorize the polynomial (a quintic, since zero is obvious), perhaps by "guessing

that -1 is a root, and (2) you must still make the partial fractions stuff, which seems far from being that easy, as \frac{1}{x(x+1)(x^2+x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{Cx+D}{x^2+x+1}+\frac{Ex+F}{(x^2+x+1)^2}
A matter of taste, I guess...perhaps because I'm a theoretical mathematician I wouldn't dare call the above "easy", or perhaps I would

but I'd add immediately "annoying and long" after that.

DonAntonio

 

[LikeMath]

Yes you are right, "easy" was not appropriate.

 

[JJacquelin]

Hi !
Who said "it's not easy" ?
It takes less time to compute it that to type it.

 

[DonAntonio]

Hi !
Who said "it's not easy" ?
It takes less time to compute it that to type it.

Really...?! Common, it's nice to show off but not with this petty things.

DonAntonio