Question

Evaluate : `int_1^2 1/((x+1)(x+3)) dx` 

Answer 1

`int_1^2 1/((x+1)(x+3)) dx`   

= `1/2int_1^2(1/(x+1)-1/(x+3))dx` 

=`1/2[{log(x+1)}_1^2-{log(x+3)}_1^2]` 

=`1/2[(log)3-log2)-(log 5-log4)]` 

=`1/2(log3+log4)-(log2+log5)` 

=`1/2[log12-log10]` 

=`(1/2)log (12/10)` 

∴`int_1^2 1/((x+1)(x+3))dx=1/2 log (6/5)`