u-substitution and IBP Practice Problems

What should $u$ equal in this indefinite integral?

 $\int\frac{\sin\left(\ln x\right)}{x}dx$  

•  $u=x$  
  

•  $u=\frac{\ln x}{x}$  
  

•  $u\ =\ \ln x$  
  

•  $u\ =\ \sin\left(\ln x\right)$  
  

Why do we need to put a $+\ C$  along with our answer to an indefinite integral? What significance does it have?

 $\int\sqrt{5x-1}\ dx$  

•  $\frac{1}{2\sqrt{5x-1}}+C$  

•  $\frac{5}{2\sqrt{5x-1}}+C$  

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

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

 $\int6x\left(x^2+1\right)^2\ dx$ 

•  $\left(x^2+1\right)^3+C$  

•  $3\left(x^2+1\right)^3+C$  

•  $\left(x^2+1\right)^2+C$  
  

•  $6\left(x^2+1\right)^2\ +C$  
  

What should $u$ equal in this integral? 

 $\int\cos x\cdot\csc^2\left(\sin x\right)\ dx$  

•  $\csc x$  
  

•  $\csc^2x$  
  

•  $\sin x$  
  

•  $\cos x$  
  

 $\int x\left(4x+1\right)^{30}\ dx$  

•  $\frac{1}{16}\left(\frac{\left(4x+1\right)^{32}}{32}-\frac{\left(4x+1\right)^{31}}{31}\right)+C$  

•  $\left(4x+1\right)^{32}-\left(4x+1\right)^{31}+C$  

•  $\frac{1}{2}\left(4x^2+1\right)^{31}+C$  

• This function does not have an antiderivative

 $\int\frac{4x}{x^2+6}dx$  

•  $\ln\left(2x^2+6\right)\ +C$  
  

•  $4x\ln\left(2x^2+6\right)\ +C$  
  

•  $\frac{1}{2x^2+6}+C$  
  

•  $\frac{2x^2}{\frac{1}{3}x^3+6x}\ +C$  
  

What should $u$ equal in this integral?

 $\int\frac{\sin\left(2x\right)}{\cos^4\left(2x\right)}dx$ 

•  $\sin\left(2x\right)$ 
  

•  $\cos^4\left(2x\right)$  
  

•  $\cos\left(2x\right)$  
  

•  $\tan^4\left(2x\right)$  

 $\int\cot x\ dx$  
(hint: write  $\cot x$  as a fraction, then perform  $u$ -substitution) 

•  $\ln\left(\sin x\right)+C$  

•  $-\ln\left(\sin x\right)+C$  

•  $\ln\left(\cos x\right)\ +C$  

•  $-\ln\left(\cos x\right)+C$  

Tags

CCSS.HSF.TF.B.7

 $\int x\ dx$  

•  $1+C$  

•  $x^2+C$  

•  $\frac{x^2}{2}+C$  

• This function does not have an antiderivative