How do you find the integral of #int 4/(1+9x^2)dx#?

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Answer 1 · 2018-03-18T20:27:51

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

#int4/(1+9x^2)dx#

substitute

#9x^2=tan^2u#

#=>3x=tanu--(1)#

#=>3dx=sec^2udu#

#int4/(1+9x^2)dx=4int1/(1+tan^2u)xx(sec^2udu)/3#

#=4/3intcancel((sec^2u)/(1+tan^2u))du#

#I=4/3intdu=4/3u+c#

#(1)rarru=tan^(-1)(3x)#

#:.I=4/3tan^(-1)(3x)+c#