How do I find the integral ∫1(w−4)(w+1)dw ?

1 Answer
Gaurav
Jul 28, 2014

=15ln(w−4w+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)=Aw−4+Bw+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=−15

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

Now,

1(w−4)(w+1)=15(w−4)−15(w+1)

Integrating both side with respect to w,

∫1(w−4)(w+1)dw=∫15(w−4)dw−∫15(w+1)dw

=15(ln(w−4)−ln(w+1))+c, where c is a constant

=15ln(w−4w+1)+c, where c is a constant

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