#intx^2/sqrt(4x+25)dx #
#Let x = 5/2 tan A, dx = (5/2 sec^2a)dA#
#=intx^2/sqrt(4x+25)dx = int (5/2)^2(tan^2A sec^2A) /(sqrt(4(5/2)^2tan^2A+25)##dA#
#=intx^2/sqrt(4x+25)dx = int (5/2)^2(tan^2A sec^2A) /(sqrt(25tan^2A+25)##dA#
#=intx^2/sqrt(4x+25)dx = int (5/2)^2(tan^2A sec^2A) /5(sqrt(tan^2A+1))##dA#
#sec^2A-1= tan^2A#
# = int (5/4)(tan^2A sec^2A) /(sqrt(sec^2A))##dA#
# = int (5/4)((sec^2A-1) sec^2A) /(sqrt(sec^2A))##dA#
# = int (5/4)((sec^3A-secA)) ##dA#
# = (5/4)(int(sec^3A)-int(secA)) ##dA# -------- equation (1)
We need to find
#int(secA)# #dA# = #int sec A(secA + tanA)/(secA+tanA)##dA#
=#int sec A(sec^2A + tanAsecA)/(secA+tanA)##dA#
Let #B= secA+tanA#
#dB##= (secAtanA+sec^2A)# #dA#
#=int1/B##dB#
#=ln |B|+c#
#=ln |secA+tanA|+c# -----equation 2
We now need to find
#int(sec^3A)# #dA#
Integrate by parts
#int sec^2AsecA# #dA#
Let #C= secA#
#dC##= secAtanA##dA#
#dD##= sec^2A# #dA#
#D= int sec^2A# #dA# #=tanA + const#
Using by parts gives
#=int sec^2AsecA# #dA# # = CD - int D##dC#/#dA#
#=secAtanA - int secAtanAtanA# #dA#
#=secAtanA - int (sec^2A-1)secA# #dA#
#=secAtanA - int (sec^3A-secA)# #dA#
#int sec^3AsecA# #dA##=secAtanA - (int sec^3A -int secA)# #dA#
substituting using equation 2
#2int sec^3A# #dA# = # secAtanA+ ln|secA+tanA)#
# 1/2(secAtanA+ ln|secA+tanA))# ----equation 3
From equation 1 we had
#intx^2/sqrt(4x+25)dx # # = (5/4)(int(sec^3A)-int(secA)) ##dA#
using the results from equations 2 and 3
#=5/4((1/2(secAtanA + ln|(SecA+tanA)|))-ln|(SecA+tanA)|) #
#=5/8(secAtanA + -ln|(SecA+tanA)|)) #
Now convert everything back to x
#x=5/2 tan A#, #tan A=2/5x#, #tan^2A=(2/5)^2x^2#
#sec^2A = tan^2A +1 =(4x^2+25)/25#
#secA = =sqrt((4x^2+25)/(25))= sqrt(4x+25)/5#
#1/20(sqrt(4x+25))x - 5/8ln(sqrt((4x+25))/(5)+2/5x)#
