How do I find the integral 1(w4)(w+1)dw ?

1 Answer
Jul 28, 2014

=15ln(w4w+1)+c, where c is a constant

Explanation :

This type of question usually solve by using Partial Fractions,

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

1(w4)(w+1)=Aw4+Bw+1

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

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

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

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

A+B=0 A=B ...........(i)
A4B=1 ..............(ii)

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

5B=1 B=15

from B, we can easily calculate A, which will be 15

Now,

1(w4)(w+1)=15(w4)15(w+1)

Integrating both side with respect to w,

1(w4)(w+1)dw=15(w4)dw15(w+1)dw

=15(ln(w4)ln(w+1))+c, where c is a constant

=15ln(w4w+1)+c, where c is a constant