Question

How do I find the integral `int1/((w-4)(w+1))dw` ?

Answer 1

`=1/5ln((w-4)/(w+1))+c`, where `c` is a constant

Explanation :

This type of question usually solve by using Partial Fractions,

`1/((w-4)(w+1))`, it can be written as

`1/((w-4)(w+1))=A/(w-4)+B/(w+1)`

multiplying by `(w-4)(w+1)` on both sides, we get

`1=A(w+1)+B(w-4)`

`1=(A+B)w+(A-4B)`

Now comparing coefficient of `w` and constants both sides, we get

`A+B=0` `=>` `A=-B` ...........`(i)`
`A-4B=1` ..............`(ii)`

Substituting value of `A` from `(i)` to `(ii)`, we get

`-5B=1` `=>` `B=-1/5`

from `B`, we can easily calculate `A`, which will be `1/5`

Now,

`1/((w-4)(w+1))=1/(5(w-4))-1/(5(w+1))`

Integrating both side with respect to `w`,

`int1/((w-4)(w+1))dw=int1/(5(w-4))dw-int1/(5(w+1))dw`

`=1/5(ln(w-4)-ln(w+1))+c`, where `c` is a constant

`=1/5ln((w-4)/(w+1))+c`, where `c` is a constant