How can you find the antiderivative of 1/(1+3x²) ?

#int dx/(1+3x^2) = int dx/(1+(sqrt3x)^2)#

#int dx/(1+3x^2) = 1/sqrt3 int (d (sqrt3x))/(1+(sqrt3x)^2)#

#int dx/(1+3x^2) = 1/sqrt3 arctan(sqrt3x)+C#

Writing
#1/(1+(sqrt(3)*x)^2)#
we Substitute
#t=sqrt(3)x#
then we get

#dx=1/sqrt(3)dx#
and we get

#1/sqrt(3)int1/(1+t^2)dt=1/sqrt(3)*arctan(t)+C=1/sqrt(3)arctan(sqrt(3)*x)+C#

#int 1/(1+3 x^2)dx =# Let #x = 1/sqrt 3 tan theta#

#tan theta = sqrt3 x :. theta = tan^-1 sqrt3 x#

#:. dx = 1/sqrt 3 sec ^2 theta d theta :. x ^2= 1/3 tan^2 theta#

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

# =int(1/sqrt 3 sec ^2 theta d theta)/(1+3*1/3tan^2 theta)#

# =int(1/sqrt 3 sec ^2 theta d theta)/(1+tan^2 theta)#

# =int1/sqrt 3 cancel(sec ^2 theta)/cancel(sec ^2 theta)*d theta#

#=1/sqrt3 theta+ C = 1/sqrt3 tan^-1 sqrt3 x+ C# ; [Ans]

#int1/(1+3x^2)dx#

Integration by Trigonometric Substitution

#x=1/sqrt(3)tanu#

#dx=1/sqrt(3)tanusecu*du#

Substitute

#int1/(1+3x^2)dx=1/sqrt3int(secutanu*du)/(1+tan^2u)#

#color(green)(1+tan^2u=sec^2u#

=#1/sqrt3int(cancel(secu)tanudu)/sec^cancel(2)u#

=#1/sqrt3intsinu*du#

=#-1/sqrt3cosu+C#

reverse the substitution

#=-1/sqrt3*1/sqrt(1+3x^2)+C#