u-substitution and IBP Practice Problems

 $u=x$  

 $u=\frac{\ln x}{x}$  

 $u\ =\ \ln x$  

 $u\ =\ \sin\left(\ln x\right)$  

 $\frac{1}{2\sqrt{5x-1}}+C$  

 $\frac{5}{2\sqrt{5x-1}}+C$  

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

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

 $\left(x^2+1\right)^3+C$  

 $3\left(x^2+1\right)^3+C$  

 $\left(x^2+1\right)^2+C$  

 $6\left(x^2+1\right)^2\ +C$  

 $\csc x$  

 $\csc^2x$  

 $\sin x$  

 $\cos x$  

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

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

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

This function does not have an antiderivative

 $\ln\left(2x^2+6\right)\ +C$  

 $4x\ln\left(2x^2+6\right)\ +C$  

 $\frac{1}{2x^2+6}+C$  

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

 $\sin\left(2x\right)$ 

 $\cos^4\left(2x\right)$  

 $\cos\left(2x\right)$  

 $\tan^4\left(2x\right)$  

 $\ln\left(\sin x\right)+C$  

 $-\ln\left(\sin x\right)+C$  

 $\ln\left(\cos x\right)\ +C$  

 $-\ln\left(\cos x\right)+C$  

 $1+C$  

 $x^2+C$  

 $\frac{x^2}{2}+C$  

This function does not have an antiderivative