a$.$ Evaluate the following integral using substitution. Submit your work. Your solution will be graded on correctness and completeness.

${\displaystyle \underset{0}{\overset{1}{}}}(5x+1{)}^{4}dx$

b$.$ Four students were asked to integrate ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^2(3-10x^3{)}^{4}dx$ using the substitution $u=3-10x^3.$ Determine which solution is correct, and explain why the others are not.

Student A

$u=3-10x^3,-30{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{u}^{4}du=-30(\frac{u^5}{5})+C=-6(3-10x^3{)}^5+C$

$\frac{du}{dx}=-30x^2$

Student B

$u=3-10x^3,-\frac{1}{30}{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{u}^{4}du=(-\frac{1}{30})(\frac{u^5}{5})+C=(-\frac{1}{150})(3-10x^3{)}^5+C$

$\frac{du}{dx}=-30x^2$

Student C

$u=3-10x^3,-\frac{1}{30}{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{u}^{4}du=-\frac{1}{30}(4u^3)+C=-\frac{4}{30}(3-10x^3{)}^3+C$

$\frac{du}{dx}=-30x^2$

Student D

$u=3-10x^3,-\frac{1}{30}{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{u}^{4}du=-\frac{1}{30}(3-10x^3{)}^4+C$

$\frac{du}{dx}=-30x^2$