# Indefinite Integration

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?)

## 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.

Homework Helper
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?

Homework Helper
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

1 person
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.

Homework Helper
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!

Homework Helper
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.

Homework Helper
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.