How do I find the integral int1/((w-4)(w+1))dw∫1(w−4)(w+1)dw ?
1 Answer
=1/5ln((w-4)/(w+1))+c=15ln(w−4w+1)+c , wherecc is a constant
Explanation :
This type of question usually solve by using Partial Fractions,
1/((w-4)(w+1))1(w−4)(w+1) , it can be written as
1/((w-4)(w+1))=A/(w-4)+B/(w+1)1(w−4)(w+1)=Aw−4+Bw+1
multiplying by
1=A(w+1)+B(w-4)1=A(w+1)+B(w−4)
1=(A+B)w+(A-4B)1=(A+B)w+(A−4B)
Now comparing coefficient of
A+B=0A+B=0 =>⇒ A=-BA=−B ...........(i)(i)
A-4B=1A−4B=1 ..............(ii)(ii)
Substituting value of
-5B=1−5B=1 =>⇒ B=-1/5B=−15
from
Now,
1/((w-4)(w+1))=1/(5(w-4))-1/(5(w+1))1(w−4)(w+1)=15(w−4)−15(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(w−4)(w+1)dw=∫15(w−4)dw−∫15(w+1)dw
=1/5(ln(w-4)-ln(w+1))+c=15(ln(w−4)−ln(w+1))+c , wherecc is a constant
=1/5ln((w-4)/(w+1))+c=15ln(w−4w+1)+c , wherecc is a constant
