simple integration question?

Question

can you integrate dx/((x+a)(x+b))

Answer 1

yes,
using partial fractions, if a and b are different:
1 /((x+a)(x+b)) = A/(x+a) + B/(x+b), let us find A and B in terms of a and b:

= ( A (x+b) + B(x+a) )/ (x+a)(x+b)
=( x(A+B) + Ab+ Ba) / (x+a)(x+b) whicho should be equal to: 1 /((x+a)(x+b))
so A+B =0 and Ab+Ba=1
A=-B, so (-B)b+Ba=1
B( a-b) =1
B= 1/(a-b) and => A= 1/(b-a)

integral dx/((x+a)(x+b))
= integral ( 1/(b-a) 1/(x+a) )dx +integral ( 1/(a-b) 1/(x+b) )dx
= ln(x+a)/(b-a) + ln(x+b) / (a-b) + C
=(1/(a-b) ) ( -ln(x+a) + ln(x+b) ) + C
=(1/(a-b) ) ln( (x+b)/(x+a) ) + C.


if a=b, then it is easier, use substitution.

Answer 2

You can integrate above integral by partial fraction

let 1/(x+a)(x+b) = A/(x+a) +B/(x+b) ......(1)

solving weget A = 1/(b-a)
B = -1/(b -a)

integrating equation 1 we get 1/(b-a) ln [(x+a)/(x+b) ]+c
where c is the constant of integration

Answer 3

try to write 1/(x+a)(x+b) as p/(x+a) + q/(x+b)

1= p*(x+b) +q*(x+a) So q= 1/(a-b) and p = 1/(b-a)

the integrand now is 1/(b-a) * ( 1/(x+a) - 1/(x+b) )

so the integral is 1/(b-a) *( logIx+aI- logIx+bI)

= 1/(b-a) *log I (x+a)/x+b)I log is in base "e"