Yes, partial fractions is the hard way- but it can be done.
so that
Taking x= 1, 5= 3A so A= 5/3. Taking x= 0, 0= (5/3)(1)+ C(-1) and C= -5/3.
Caution: it should be
Tonio
Taking x= -1, 5= 5/3+ (-5/3- B)(-2)= 2B+ 15/3= 2B+ 5 so B= 0.
Your integral is
The first is easy to integrate. To integrate the second, complete the square:
so the integral is
[tex]-\frac{5}{3}\int \frac{dx}{(x+1/2)^2+ 3/4}[/itex]
and the substitution u= x+ 1/2 makes that
[tex]-\frac{5}{3}\int \frac{du}{u^2+ 3/4}= -\frac{20}{9}\int\frac{du}{(4u^2}/3+ 1}[/itex]
and the final substitution
makes it