Solving the Indefinite Integration Problem

[Saitama]

Homework Statement

The value of \int_0^{1} (\prod_{r=1}^{n} (x+r))(\sum_{k=1}^{n} \frac{1}{x+k}) dx equals:
a)n
b)n!
c)(n+1)!
d)n.n!

(Can someone tell me how to make bigger parentheses using latex?)

Homework Equations

The Attempt at a Solution

I know that the question becomes a lot easier if i put n=1 or 2 and then integrate. But i was wondering if there is any proper way to solve it. I can't go further after expanding the given expression.
\int_{0}^{1} (x+1)(x+2)...(x+n)(\frac{1}{x+1}+\frac{1}{x+2}+...\frac{1}{x+n})dx
which is equal to
\int_{0}^{1} \sum_{r=1}^n \frac{(x+n)!}{x+r}
I am stuck now, i can't find any way further.

Any help is appreciated.

 

[Saitama]

I still don't have any clue.

 

[ehild]

What is the derivative of a product? What is the derivative of (x+1)(x+2)? and (x+1)[(x+2)(x+3)] ?

ehild

 

[Saitama]

What is the derivative of a product? What is the derivative of (x+1)(x+2)? and (x+1)[(x+2)(x+3)] ?

ehild

Derivative of (x+1)(x+2)=2x+3
Derivative of (x+1)(x+2)(x+3)=3x²+12x+11

I still don't get any idea how this would help?

 

[ehild]

Do not simplify.

d/dx[(x+1)(x+2)]=(x+2)+(x+1)=(x+1)(x+2)/(x+1)+(x+1)(x+2)/(x+2)=[(x+1)(x+2)][1/(x+1)+1/(x+2)]

d/dx[(x+1)(x+2)(x+3)]=(x+2)(x+3)+(x+1)(x+3)+(x+1)(x+2)=[(x+1)(x+2)(x+3)][1/(x+1)+1/(x+2)+1(x+3)]

Do you see?

ehild

 

[Saitama]

Wait, i guess i have got it, i will be back in a few hours with a solution.
Thanks for the help!

EDIT: Oops, seems like you posted just a few seconds before me.

 

[ehild]

Wait, i guess i have got it, i will be back in a few hours with a solution.
Thanks for the help!

EDIT: Oops, seems like you posted just a few seconds before me.

Well, I looking forward to the solution

ehild

 

[Saitama]

So here's the solution:
Let
t=(x+1)(x+2)(x+3)...(x+n)
\ln t=\ln(x+1)+\ln(x+2)+\ln(x+3)...+\ln(x+n)
\frac{1}{t}\frac{dt}{dx}=\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}...\frac{1}{x+n}
dt=(x+1)(x+2)(x+3)...(x+n)(\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}...+\frac{1}{x+n})dx
Therefore the original question becomes ∫dt, upper limit is (n+1).n! and lower limit is n!.
Solving, i get the answer as d) option.

Thanks for the help!

 

[Mentallic]

By the way, since it seems like your little sub-question about the larger brackets was missed, I'll chime in on that.

I myself use /left( and /right) which looks like: \left(\frac{1}{1}\right) but there are other more specific "fonts" you can use, which ranges from smallest to largest:

/big( \big(\frac{1}{1}\big)
/Big( \Big(\frac{1}{1}\Big)
/bigg( \bigg(\frac{1}{1}\bigg)
/Bigg( \Bigg(\frac{1}{1}\Bigg)

And there could be others, but that should about cover it.

 

[ehild]

So here's the solution:
Let
t=(x+1)(x+2)(x+3)...(x+n)
\ln t=\ln(x+1)+\ln(x+2)+\ln(x+3)...+\ln(x+n)

\frac{1}{t}\frac{dt}{dx}=\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}...\frac{1}{x+n}
dt=(x+1)(x+2)(x+3)...(x+n)(\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}...+\frac{1}{x+n})dx

That is ingenious! You are really cool

Therefore the original question becomes ∫dt, upper limit is (n+1).n! and lower limit is n!.
Solving, i get the answer as d) option.

Thanks for the help!

Very good. You are better and better every day!

ehild

 

[Saitama]

By the way, since it seems like your little sub-question about the larger brackets was missed, I'll chime in on that.

I myself use /left( and /right) which looks like: \left(\frac{1}{1}\right) but there are other more specific "fonts" you can use, which ranges from smallest to largest:

/big( \big(\frac{1}{1}\big)
/Big( \Big(\frac{1}{1}\Big)
/bigg( \bigg(\frac{1}{1}\bigg)
/Bigg( \Bigg(\frac{1}{1}\Bigg)

And there could be others, but that should about cover it.

Thanks Mentallic, really helpful.

That is ingenious! You are really cool

I did nothing, you almost solved it by giving me a simple hint. Thanks!

 

[ehild]

That method of using logarithm was entirely your idea . I did not think of it.

ehild