The indefinite integral: `int (dw)/w^3`



Asked on

2 Answers | Add Yours

justaguide's profile pic

Posted on (Answer #1)

The integral `int (dw)/w^3` has to be determined.

`int (dw)/w^3`

=> `int w^-3 dw`

=> `(w^(-3+1))/(-3+1)`

=> `w^-2/-2`

=> `-1/(2*w^2)`

The integral `int (dw)/w^3 = -1/(2*w^2) + C`

embizze's profile pic

Posted on (Answer #2)

Determine `int (dw)/w^2` :

`int (dw)/(w^2)=int w^(-2)dw` Use the general power rule:


`=-1/w +C`

Thus `int (dw)/(w^2)=-1/w + C`


Check: if we take the derivative we should get the integrand back:

`d/(dw)( -1/w +C)=d/(dw)(-w^(-1))+d/(dw)(C)`


`=1/w^2` as required.

We’ve answered 396,523 questions. We can answer yours, too.

Ask a question