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Answer 1 · 2017-08-31T20:09:46

#intx^2/(9+16x^6)dx = 1/36arctan(4/3x^3)+"constant"#

For the integral #F(x)=intx^2/(9+16x^6)dx#, sub #u=x^3 # and #du=3x^2dx#

Then #F(x) = 1/3int1/(9+16u^2)du = 1/27int1/(1+16/9u^2)du#

For the integral #F(x)#, sub #t=4/3u# and #dt=4/3du#

Then #F(x) = 1/36int1/(1+t^2)dt#

#F(x)# is now in a standard form and can be evaluated as a standard integral.

#F(x)=1/36int1/(1+t^2)dt=1/36arctant = 1/36arctan(4/3u) = 1/36arctan(4/3x^3)+"constant"#

Answer 2 · 2017-09-01T16:15:45

#1/36tan^-1(4/3x^3)+C#

#int(x^2dx)/(9+16x^6)#

Use the substitution #x^3=3/4tantheta,=>3x^2dx=3/4sec^2thetad theta#.

#=1/3int(3x^2dx)/(9+16(x^3)^2)=1/3int(3/4sec^2thetad theta)/(9+16(9/16tan^2theta))#

#=1/4int(sec^2thetad theta)/(9(1+tan^2theta))#

Recall that #1+tan^2theta=sec^2theta#:

#=1/36intd theta=1/36theta+C#

From #x^3=3/4tantheta#, note that #tantheta=4/3x^3# so #theta=tan^-1(4/3x^3)#:

#=1/36tan^-1(4/3x^3)+C#