Integral by parts

What would you choose for your u here if you used integration by parts?

• t

• 3t

• e^2t

• e^t

• don't use IBP, let u = 2t

Evaluate the indefinite integral using integration by parts. 
 $\int3x\ e^{2x}\ dx\$  

•  $-\frac{xe^{2x}}{2}+\frac{e^{2x}}{4}+C$  

•  $\frac{3xe^{2x}}{2}-\frac{3e^{2x}}{4}+C$  

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

•  $-\frac{xe^{2x}}{2}+\frac{\ln e^{2x}}{4}+C$  

Evaluate the indefinite integral using integration by parts. 
 $\int t^2\ln t\ dt\$  

•  $\frac{2t^2\ln2t-t^2}{4}+C$  

•  $\frac{t^3\ \ln3}{3}-\frac{t^3}{9}+C$  

•  $\frac{e^t}{2t+2}+C$  

•  $\frac{-2t-1}{4e^{2t}}+C$  

 $\int xe^{-x}dx$  

•  $=-e^{-x}\left(x-1\right)+C$  

•  $=e^{-x}\left(x+1\right)+C$  

•  $=e^{-x}\left(x-1\right)+C$  

•  $=-e^{-x}\left(x+1\right)+C$  

Evaluate the indefinite integral using integration by parts
 $\int xe^{4x}dx\ ;\ \$  

•  $-\frac{x}{4^x\ln4}-\frac{1}{4^x\cdot\left(\ln4\right)^2}+C$  

•  $\frac{x^5\ln x}{5}-\frac{x^5}{25}+C$  

•  $x\ln\left(x+4\right)-x+4\ln\left(x+4\right)+C$  

•  $\frac{xe^{4x}}{4}-\frac{e^{4x}}{16}+C$  

Evaluate the indefinite integral using integration by parts. 
 $\int3x\ e^{2x}\ dx\$  

•  $-\frac{xe^{2x}}{2}+\frac{e^{2x}}{4}+C$  

•  $\frac{3xe^{2x}}{2}-\frac{3e^{2x}}{4}+C$  

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

•  $-\frac{xe^{2x}}{2}+\frac{\ln e^{2x}}{4}+C$  

Evaluate the indefinite integral using integration by parts. u and dv are provided.
 $\int x^{4\ }\ln x\ dx\ ;\ \ u=\ln x,\ dv=x^4dx$  

•  $\frac{e^x}{4x+4}+C$  

•  $\frac{2x^{\frac{3}{2}}\ln4x}{3}-\frac{4x^{\frac{3}{2}}}{9}+C$  

•  $\frac{\left(4x^2-1\right)\cdot e^{4x^2}}{32}+C$  

•  $\frac{x^5\ln x}{5}-\frac{x^5}{25}+C$  

Evaluate the indefinite integral using integration by parts. 
 $\int3x\ e^{2x}\ dx\$  

•  $-\frac{xe^{2x}}{2}+\frac{e^{2x}}{4}+C$  

•  $\frac{3xe^{2x}}{2}-\frac{3e^{2x}}{4}+C$  

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

•  $-\frac{xe^{2x}}{2}+\frac{\ln e^{2x}}{4}+C$  

Evaluate the indefinite integral using integration by parts. u and dv are provided.
 $\int xe^{4x}dx\ ;\ \ \ u=x,\ dv=e^{4x}dx$  

•  $-\frac{x}{4^x\ln4}-\frac{1}{4^x\cdot\left(\ln4\right)^2}+C$  

•  $\frac{x^5\ln x}{5}-\frac{x^5}{25}+C$  

•  $x\ln\left(x+4\right)-x+4\ln\left(x+4\right)+C$  

•  $\frac{xe^{4x}}{4}-\frac{e^{4x}}{16}+C$  

Evaluate the indefinite integral using integration by parts. u and dv are provided.
 $\int x^{4\ }\ln x\ dx\ ;\ \ u=\ln x,\ dv=x^4dx$  

•  $\frac{e^x}{4x+4}+C$  

•  $\frac{2x^{\frac{3}{2}}\ln4x}{3}-\frac{4x^{\frac{3}{2}}}{9}+C$  

•  $\frac{\left(4x^2-1\right)\cdot e^{4x^2}}{32}+C$  

•  $\frac{x^5\ln x}{5}-\frac{x^5}{25}+C$