Answer
$${\text{part a) }}\frac{1}{4}\tan \left( {4x + 1} \right) + C{\text{ part b)}}\frac{1}{6}{\left( {1 + 2{y^2}} \right)^{3/2}} + C$$
Work Step by Step
$$\eqalign{
& {\text{part a}} \cr
& \cr
& \int {{{\sec }^2}\left( {4x + 1} \right)} dx \cr
& {\text{substitute }}u = 4x + 1,{\text{ }}du = 2dx \cr
& = \int {{{\sec }^2}\left( {4x + 1} \right)} \left( {\frac{1}{4}} \right)du \cr
& = \frac{1}{4}\int {{{\sec }^2}u} du \cr
& {\text{find antiderivative}} \cr
& = \frac{1}{4}\tan x + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{4}\tan \left( {4x + 1} \right) + C \cr
& \cr
& {\text{part b}} \cr
& \int {y\sqrt {1 + 2{y^2}} } dy \cr
& {\text{substitute }}u = 1 + 2{y^2},{\text{ }}du = 4ydy \cr
& = \int {\sqrt u } \left( {\frac{1}{4}} \right)du \cr
& = \frac{1}{4}\int {{u^{1/2}}} du \cr
& {\text{find antiderivative}} \cr
& = \frac{1}{4}\left( {\frac{{{u^{3/2}}}}{{3/2}}} \right) + C \cr
& = \frac{1}{6}{u^{3/2}} + C \cr
& {\text{write in terms of }}y \cr
& = \frac{1}{6}{\left( {1 + 2{y^2}} \right)^{3/2}} + C \cr} $$
