Review Integration

 $\int\cos\left(4x+5\right)dx_{ }^{ }$   

•  $-\frac{1}{4}\sin\left(4x+5\right)+C$  

•  $4\sin\left(4x+5\right)+C$  

•  $\frac{1}{4}\sin\left(4x+5\right)+C$  

•  $4\cos\left(4x+5\right)+C$  

• 3

• ½

• -3/2

• 4/11

 $\int_{ }^{ }25\sec^2xdx$  

• 25 secx tanx + C

• 25 tanx + C

• -25 tanx + C

• -25 sec x + C

If

 $h'\left(x\right)=x\left(3x^{2\ }+2\right)^{10}$  
then  $h\left(x\right)=$  

•  $\frac{1}{11}\left(3x^{2\ }+2\right)^{11}+C$  

•  $\frac{1}{6}\left(3x^{2\ }+2\right)^{11}+C$  

•  $\frac{6}{11}\left(3x^2+2\right)_{ }^{11}+C$  

•  $\frac{1}{66}\left(3x^{2\ }+2\right)^{11}+C$  

 $\int_{ }^{ }\left(3x^4+2\right)^4\left(12x^3\right)dx$  

•   $\frac{1}{2}\left(3x^4+2\right)^4+C$  

•  $\left(3x^4+2\right)^4+C$  

•  $\frac{1}{5}\left(3x^{4\ }+2\right)^{5\ }+C$  

•  $\frac{1}{15}\left(3x^4+2\right)^5+C$  

• A

• B

• C

• D

What should "u" equal in this integral?

• $\tan x\sec x\ dx$
  

• $\sec x\ dx$

• $\sec^2x\ dx$

• $\tan x\ dx$
  

 $\int_1^6\left(\frac{x}{\sqrt{x+2}}\right)dx$  

If the above integral is to be solved with u-substitution, which answer choice is the proper set-up for the u-substitution?

•  $\int_3^8\ \ u^{-\frac{1}{2}}\left(u-2\right)du$  

•  $\int_1^6\ \ u^{-\frac{1}{2}}\left(u-2\right)du$  

•  $\int_1^6\ \ u^{-\frac{1}{2}}du$  

•  $\int_3^8\ \ u^{-\frac{1}{2}}du$  

Which integral below needs to be solved with u-sub?

• $\int_{ }^{ }\sin xdx$

• $\int_{ }^{ }\sec^2xdx$

• $\int_{ }^{ }\sqrt{2x-1}dx$

• $\int_{ }^{ }\left(\frac{1}{x^2}\right)dx$

 $\int_0^3\ \sqrt{9-x^2}dx$  

•  $\frac{9\pi}{4}$  

•  $\frac{9\pi}{2}$  

•  $4.5$  

•  $13.5$