How do I find the integral int1/((w-4)(w+1))dw1(w4)(w+1)dw ?

1 Answer
Jul 28, 2014

=1/5ln((w-4)/(w+1))+c=15ln(w4w+1)+c, where cc is a constant

Explanation :

This type of question usually solve by using Partial Fractions,

1/((w-4)(w+1))1(w4)(w+1), it can be written as

1/((w-4)(w+1))=A/(w-4)+B/(w+1)1(w4)(w+1)=Aw4+Bw+1

multiplying by (w-4)(w+1)(w4)(w+1) on both sides, we get

1=A(w+1)+B(w-4)1=A(w+1)+B(w4)

1=(A+B)w+(A-4B)1=(A+B)w+(A4B)

Now comparing coefficient of ww and constants both sides, we get

A+B=0A+B=0 => A=-BA=B ...........(i)(i)
A-4B=1A4B=1 ..............(ii)(ii)

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

-5B=15B=1 => B=-1/5B=15

from BB, we can easily calculate AA, which will be 1/515

Now,

1/((w-4)(w+1))=1/(5(w-4))-1/(5(w+1))1(w4)(w+1)=15(w4)15(w+1)

Integrating both side with respect to ww,

int1/((w-4)(w+1))dw=int1/(5(w-4))dw-int1/(5(w+1))dw1(w4)(w+1)dw=15(w4)dw15(w+1)dw

=1/5(ln(w-4)-ln(w+1))+c=15(ln(w4)ln(w+1))+c, where cc is a constant

=1/5ln((w-4)/(w+1))+c=15ln(w4w+1)+c, where cc is a constant