Integration by Parts 2/1/2021

What is the Integration by Parts Formula?

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Tags

CCSS.HSA.SSE.A.2

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

Tags

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

CCSS.HSF.IF.A.2

CCSS.HSF.LE.A.1

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\$  

•  $\frac{2t^2\ln2t-t^2}{4}+C$  

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

•  $\frac{e^t}{2t+2}+C$  

•  $\frac{-2t-1}{4e^{2t}}+C$  

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

•  $\frac{e^x}{4x+4}+C$  

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

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

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

 $\int_{ }xe^x\ dx$  by using integration by parts.

•  $xe^x-\ e^x+c$  

•  $xe^x+e^x+c$  

•  $xe^x+c$  

•  $xe^x-e^x$  

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

Tags

CCSS.HSF.TF.A.3

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

• x
• x⁴
• e^-x
• e^x

 $\int x\sin\left(8x\right)dx$  

•  $=-\frac{x}{8}\cos\left(8x\right)+\frac{1}{64}\sin\left(8x\right)+C$  

•  $=-\frac{x}{8}\cos\left(8x\right)-\frac{1}{64}\sin\left(8x\right)+C$  

•  $=-\frac{x}{8}\cos\left(8x\right)+\frac{1}{8}\sin\left(8x\right)+C$  

•  $=\frac{x}{8}\cos\left(8x\right)+\frac{1}{64}\sin\left(8x\right)+C$