Strategies for Solving Fractional Integrals with Polynomial Expressions

[Penultimate]

Can you help me solve this one :

$$\int$$ $$\frac{x+8}{x-8}$$dx

 

[Hootenanny]

Can you help me solve this one :

$$\int$$ $$\frac{x+8}{x-8}$$dx

What have you tried thus far?

 

[qntty]

Do you know how to do $\int{\log{x} \, dx}$?

 

[JG89]

(x+8)/(x-8) = (x + 8 - 16 + 16)/(x-8) = (x-8 + 16)/(x-8) = 1 + 16/(x-8).

That's not too hard to integrate, right?

 

[Penultimate]

Sorry guys i have submited the integral without puting the exponent for each X(thats why this one is too simple).

The one i am trying to solve is :

$$\int$$ $$\frac{x^3+8}{x^3-8}$$dx

 

[Penultimate]

I have tryed irracional mode (thats the one it requires but i am stuck at the end).

 

[Hootenanny]

I have tryed irracional mode (thats the one it requires but i am stuck at the end).

Why don't you show us what you've done and where your stuck then we can help you?

 

[physicsnoob93]

Hi Penultimate!

Alright, let's help you out here. Firstly, can you break up the denominator into a linear factor multiplied by a quadratic expression?

 

[JG89]

Well (x^3 + 8)/(x^3 - 8) = (x^3 - 8 + 16)/(x^3 - 8) = 1 + 16/(x^3 - 8).

x^3-8 can easily be factored, and so use partial fractions on that integral.

 

[HallsofIvy]

Actually, $x^3+ 8$ can be factored as easily as $x^3- 8$. In fact, [itexs]x^n+ a^n[/itex] can be factored easily for all odd n.

 

[Дьявол]

$$x^3+8=x^3+2^3=(x+2)(x^2-2x+4)$$

$$x^3-8=x^3-2^3=(x-2)(x^2+2x+4)=(x-2)(x+2)^2$$

Another way is:
$$\frac{x^3+8}{x^3-8}=\frac{x^3-8+16}{x^3-8}=1+\frac{16}{x^3-8}$$

Regards.