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

1 Answer
Jul 28, 2014

=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