How do I find the integral int1/((w-4)(w+1))dw ?
1 Answer
Jul 28, 2014
=1/5ln((w-4)/(w+1))+c , wherec 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
1=A(w+1)+B(w-4)
1=(A+B)w+(A-4B)
Now comparing coefficient of
A+B=0 => A=-B ...........(i)
A-4B=1 ..............(ii)
Substituting value of
-5B=1 => B=-1/5
from
Now,
1/((w-4)(w+1))=1/(5(w-4))-1/(5(w+1))
Integrating both side with respect to
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 , wherec is a constant
=1/5ln((w-4)/(w+1))+c , wherec is a constant