How do you integrate #int dx/(4x^2+9)^2# using trig substitutions?

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Answer 1 · 2018-03-21T02:16:02

#int dx/(4x^2+9)^2=1/108arctan((2x)/3)+x/(72x^2+162)+C#

#int dx/(4x^2+9)^2#

=#1/2int (2dx)/((2x)^2+3^2)^2#

After using #2x=3tanu# and #2dx=3(secu)^2*du# transforms, this integral became

#1/2int (3(secu)^2*du)/(81(secu)^4)#

=#1/54int (cosu)^2*du#

=#1/108int (1+cos2u)*du#

=#1/108u+1/216sin2u+C#

=#1/108u+1/216*(2tanu)/((tanu)^2+1)+C#

After using #2x=3tanu#, #tanu=(2x)/3# and #u=arctan((2x)/3)# inverse transforms, I found

#int dx/(4x^2+9)^2#

=#1/108arctan((2x)/3)+1/216*(2*(2x)/3)/(((2x)/3)^2+1)+C#

=#1/108arctan((2x)/3)+x/(72x^2+162)+C#