# Who can solve this integral?

1. Aug 20, 2008

### tommy_ita

Can someone solve this integral as the answer, please? I'm sure it is easy for you:

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2. Aug 20, 2008

### Defennder

What have you attempted? Use integration by parts.

3. Aug 20, 2008

### tommy_ita

hi,

integration by parts seems to be right!

I just don't get through the first steps... help! ;)

4. Aug 20, 2008

### Defennder

What is formula for integration by parts? What should you let u and dv be?

5. Aug 20, 2008

### tommy_ita

S u dv = uv - S v du

u= ln(x+5)
dv= (x^3)/3

I just don't know how to solve it if where there is an addition after ln(...)

thanks

6. Aug 20, 2008

### Defennder

dv isn't x^3/3, that's v.

So you need to find du now. Or du/dx. Check out the derivative of function ln(f(x)). Then plug that into the formula.

7. Aug 20, 2008

### tommy_ita

thanks

8. Aug 20, 2008

### tommy_ita

hi, now I'm there

ln(x+5) x^3/3 - 1/3 S x^3 1/(x+5) dx

how can i solve this integral?

S x^3 1/(x+5) dx

thank you again!!!

9. Aug 20, 2008

### tiny-tim

Welcome to PF!

Hi tommy_ita! Welcome to PF!

(have an integral: ∫ and a cubed: ³)

You mean ∫x³dx /(x + 5) …

either long-division to get a constant/(x + 5) plus a quadratic,

or substitute y = x + 5, integrate, and substitute back again.

(in hindsight, making that substitution before integrating by parts might have been simpler )

10. Aug 21, 2008

### tommy_ita

hi tiny-tim,

thanks for the ∫ :D

unfortunately, I am not able to find the solution. Could you help me by doing the integral step-by-step?
that'd be very nice

∫x³dx /(x + 5) =

11. Aug 21, 2008

### tiny-tim

Hi tommy_ita!

When in doubt. use the obvious substitution, in this case:

y = x + 5, dy = dx,

∫x³dx /(x + 5) = ∫(y - 5)³dy /y = ∫(Ay² + By + C + (D/y))dy …

and you can fill in the rest yourself, can't you?