Please fully explain how to solve.

Find

${\displaystyle \underset{1}{\overset{}{}}}\frac{1}{x^7}dx$

A$.$

B$.\frac{1}{7}$

C$.1$

D$.\frac{6}{37}$

E$.\frac{1}{6}$

$7.(5$ points$)$ What is the form of the partial fraction decomposition of

$\frac{5x^2-7x1}{x^3(x^{2}7)}?$

A$.\frac{A}{x^3}\frac{BxC}{x^{2}7}$

B$.\frac{A}{x}\frac{B}{x^2}\frac{C}{x^3}\frac{DxE}{x^{2}7}$

C$.\frac{AxB}{x^3}\frac{CxD}{x^{2}7}$

D$.\frac{A}{x}\frac{B}{x^2}\frac{C}{x^3}\frac{D}{x^{2}7}$

E$.\frac{A}{x}\frac{B}{x^3}\frac{D}{x7}\frac{E}{x^{2}7}$

$8.(5$ points$)$ If $sin()=2x,$ then what is $tan()?$

A$.\frac{1}{\sqrt[\phantom{2}]{1-4x^2}}$

B$.\frac{x}{\sqrt[\phantom{2}]{4-x^2}}$

C$.\frac{\sqrt[\phantom{2}]{1-4x^2}}{2x}$

D$.\frac{2x}{\sqrt[\phantom{2}]{1-4x^2}}$

E$.\frac{\sqrt[\phantom{2}]{4-x^2}}{x}$