Review IBP and Improper Integrals

What is the Integration by Parts Formula?

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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. u and  v' are provided.
 $\int x^{4\ }\ln x\ dx\ ;\ \ u=\ln x,\ v\ '=x^4$  

•  $\frac{e^x}{4x+4}+C$  

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

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

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

Evaluate the indefinite integral using integration by parts. u and v ' are provided.
 $\int t\sin t\ dt\ ;\ \ \ u=t,\ v\ '=\sin t\$  

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

•  $-t\cos t+\sin t+C$  

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

•  $t\sin t+\cos t+C$  

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

• x

• sin(x)

• cos(x)

• e^x

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Evaluate: $\int_2^∞\frac{dx}{x^{2\ }}$   

•  $\frac{1}{2}$  

•  $\ln2$  

• 1

• 3

• divergent

Evaluate: $\int_4^∞\frac{-2x}{\left(9-x^2\right)^{\frac{1}{3}}}dx$   

• divergent

•  $\left(\frac{3}{2}\right)^{\frac{2}{3}}$  

•  $9^{\frac{2}{3}}+7^{\frac{2}{3}}$  

•  $\frac{3}{2}\left(9^{\frac{2}{3}}+7^{\frac{2}{3}}\right)$  

Evaluate: $\int_0^∞x^2e^{-x^3}dx$   

•  $-\frac{1}{3}$  

• 1

•  $\frac{1}{3}$  

• divergent