First reduce the radical to a sum of squares:
#int(sqrt(13+25x^2))dx = 1/5int(sqrt((sqrt(13)/5)^2+x^2))dx#
Pose #x=sqrt(13)/5tan(t)# and as #(d tan(t))/dt = sec^2(t)#
#dx=sqrt(13)/5sec^2(t)dt#
#int(sqrt(13+25x^2))dx = 1/5int(sqrt(13/25+13/25tan^2(t))sqrt(13)/5sec^2(t)dt = 13/125int(sqrt(1+tan^2(t))sec^2(t)dt#
Use the trigonometric identity:
#1+tan^2t=sec^2t#
#int(sqrt(13+25x^2))dx = 13/125int(sqrt(sec^2(t))sec^2(t)dt =#
#= 13/125int|sec(t)|^3dt#.
Now, to integrate this last integral let's limit ourselves in the interval #t in [-pi/2,pi/2]# where #sec(t) # is positive and use integration by parts:
#intsec^3(t)dt = int (sect*sec^2t)dt = intsect * d(tant) = sect tant -int tan t* d(sec t) = sect tant -int tan ^2t*sec t dt = #
but #tan^2t = sec^2t-1#, so
#intsec^3(t)dt = sect tant + int (sect -sec^3t)dt= sect tant +int sect dt -int sec^3(t)tdt#
or
#2intsec^3(t)dt = sect tant +int sect dt #
Finally:
#intsec^3(t)dt = 1/2sect tant +1/2 ln(|sec t + tan t|)#
Now reverse the substitution:
#tan t = 5/sqrt(13)x#
and #sect = sqrt(1+tan^2(t)) = sqrt(1+25/13x^2)#
So:
#int(sqrt(13+25x^2))dx = 5/(2sqrt(13))xsqrt(1+25/13x^2)+1/2ln(|5/sqrt(13)x+sqrt(1+25/13x^2)|)#
