Integrating Perfect Cube in Denominator: 2/(x-2)^3 dx | Homework Solution

[lost_math]

Homework Statement

\ 2/(x-2)^3 dx
Basically integrating a perfect cube in the denominator with a constant in the numerator

Homework Equations

The Attempt at a Solution

i thought it would be a form of ln(x), but then, that would mean having atleast some x terms in the numerator which are not there, so, how do i do this? Is there a known pre-fixed solution for these things? Like exp(something)?

 

[arunbg]

Use \int x^ndx = \frac{x^{n+1}}{n+1} + c

 

[cristo]

Try writing as 2(x-2)^-3. Do you know how to integrate x^-3?

 

[lost_math]

Thanks Arun and Cristo.
Cristo, I don't know how to integrate x^-3. (i think i might be hopeless, right?)
Arun, what is the expansion formula you just gave me called? Is there a name for solving by that method? It is applicable to negative powers as well?

 

[cristo]

Arunbg's formula is the formula for integrating a polynomial function of x. Thus, to integrate x^-3 you would use that formula.

 

[sara_87]

you want to integrate 2(x-2)^-3

using this \int x^n = \frac{x^{n+1}}{n+1} + c

the first thing you notice is that you add 1 to the power, then you divided by the new power
after that you multiply by the differential of what is inside the brackets

 

[sara_87]

in your question the power is -3

 

[cristo]

you want to integrate 2(x-2)^-3

using this \int x^n = \frac{x^{n+1}}{n+1} + c

the first thing you notice is that you add 1 to the power, then you divided by the new power
after that you multiply by the differential of what is inside the brackets

The last bit should read: you divide by the derivative of the term inside the brackets.

(I know it's probably a typo, and it doesn't matter in this case; but it may confuse the OP in future if left uncorrected)

 

[sara_87]

yes definitely a typo
you always divid by the differential of the inside of the brackets when integrating!