Can Calculus Substitutions Simplify Complex Integrals?

[tommy_ita]

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

see attached picture, please! :)

 

[Defennder]

What have you attempted? Use integration by parts.

 

[tommy_ita]

hi,

integration by parts seems to be right!

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

 

[Defennder]

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

 

[tommy_ita]

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

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

 

[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.

 

[tommy_ita]

thanks

 

[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!

 

[tiny-tim]

Welcome to PF!

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?

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 )

 

[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) =

 

[tiny-tim]

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) =

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?