#f(x) = int_2^x dt/((t+3)(t^2+4))#
Apply partial fractions decomposition:
#1/((t+3)(t^2+4)) = A/(t+3)+ (Bt+C)/(t^2+4)#
#1 = At^2+4A + Bt^2+Ct+3Bt+3C#
#1 = (A+B)t^2 +(C+3B)t+3C+4A#
#{(A+B=0),(C+3B=0),(3C+4A=1):}#
#{(A=-B),(-3A+C=0),(4A+3C=1):}#
#{(A=1/13),(B=-1/13),(C=3/13):}#
Then using the linearity of the integral:
#f(x) = 1/13int_2^x dt/(t+3) -1/13int_2^x (tdt)/(t^2+4) +3/13 int_2^x dt/(t^2+4) #
Solve the integrals separately:
#int_2^x dt/(t+3) = int_2^x (d(t+3))/(t+3) = [ln abs (t+3)]_2^x = ln abs(x+3)-ln5#
#int_2^x (tdt)/(t^2+4)= 1/2 int_2^x (d(t^2+4))/(t^2+4) = 1/2 [ln (t^2+4)]_2^x = ln (x^2+4)-ln 8#
#int_2^x dt/(t^2+4) = 1/2 int_2^x (d(t/2))/((t/2)^2+1) = 1/2[arctan(t/2)]_2^x = 1/2(arctan(x/2) -arctan(1)) = 1/2arctan(x/2) - pi/8#
Putting it together:
#f(x) = (2lnabs(x+3) -ln(x^2+4)+3arctan(x/2))/26 - (8ln5 -4ln8 +3pi)/104#
