Find the derivative and evaluate the integral.

Question

y=tan^-1(x²)

For this, I would utilize the derivative of arctan and chain rule, right? So with this, I'd get 1/(1+(x²)²)*(2x)? And by simplification, the answer would be (2x)/(1+(x²)²)?

∫(from 0 to √(3)/4) dx/(1+16x²)

Answer 1

derivative of arctan(x^2)

= 2x/(1+x^4)

you got it right

apply the chain rule where u=x^2, u'=2x

(arctanu)'

=1/(1+u^2) u'

=1/(1+x^2^2) (x^2)'

= 1/(1+x^4))(2x)

= 2x/(1+x^4)