Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 14

int 1/((x+a)(x+b)) dx
sol:
int 1/((x+a)(x+b)) dx
let [u= x+a] => du = dx
so ,
int 1/((x+a)(x+b)) dx
=int 1/((u)(x+b)) du
As u =x+a => x= u-a , on substituting in the above equation we get ,
=int 1/((u)(x+b)) du
=int 1/((u)((u-a)+b)) du
=int 1/((u)(u+b-a)) du
Taking partial fractions we obtain,
=int 1/((u)(u+b-a)) du
=int ((1/((a-b)(u+b-a)))+ (1/(u*(b-a)))) du
=int (1/((a-b)(u+b-a)))du+ int (1/(u*(b-a))) du -----------------(1)

Now let us consider
int (1/((a-b)(u+b-a)))du
let v=u+b-a => dv =du
so,
int (1/((a-b)(u+b-a)))du
= (1/(a-b)) int (1/(u+b-a))du
=(1/(a-b)) int (1/(v))dv
=(1/(a-b)) ln(v)
=(1/(a-b)) ln(u+b-a)
=(1/(a-b)) ln(x+b) as [u-a =x] ----------------------(2)
Now consider ,
int (1/(u*(b-a))) du
As similar to above we obtain as follows,
int (1/(u*(b-a))) du
= (1/(b-a))int (1/(u)) du
=(1/(b-a)) ln(u)
= (1/(b-a)) ln(x+a) as u=x+a -------------(3)
substituting (2) and (3) in (1) we get,
int 1/((x+a)(x+b)) dx
=int (1/((a-b)(u+b-a)))du+ int (1/(u*(b-a))) du
= (1/(a-b)) ln(x+b) +(1/(b-a)) ln(x+a) + C

is the solution :)