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Integration by Parts 

Be careful that you know the difference between integration by substitution and integration by parts. 
Integration by parts is an extremely important and useful integration technique. Most integrals you come across won't be in a simple form. The idea is to 'reduce' or alter the original integral by breaking it up into pieces that can then be evaluated using the techniques you know so far. You will find that even a small change in an integral will often make a big change in the technique used and the final solution. 

Integration by parts refers to the use of the equation \(\int{ u~dv } = uv  \int{ v~du }\).
It is important to read the next section to understand where this comes from.
Deriving the Integration By Parts Equation 

The integration by parts equation comes from the product rule for derivatives. We will show an informal proof here.
Let u and v be functions of t.
Using the product and chain rules, we take the derivative of \(u \cdot v\) to get 
\(\displaystyle{ \frac{d}{dt}[ u \cdot v] = u \frac{dv}{dt} + v \frac{du}{dt} }\) 
Now we isolate the \(\displaystyle{ u \frac{dv}{dt} }\) term to get 
\(\displaystyle{ u \frac{dv}{dt} = \frac{d}{dt}[ u \cdot v]  v \frac{du}{dt} }\) 
Next, we integrate both sides with respect to \(t\) to get 
\(\displaystyle{ \int{ u \frac{dv}{dt} ~dt } = }\) \(\displaystyle{ \int{ \frac{d}{dt}[ u \cdot v] ~dt }  \int{ v \frac{du}{dt} ~dt } }\) 
Using the idea of differentials, we end up with 
\(\displaystyle{ \int{ u ~dv } = u \cdot v  \int{ v ~du } }\) 
Here is a video showing the same derivation as above.
PatrickJMT  Deriving the Integration by Parts Formula  Easy!  
Using Integration By Parts 

The key to using integration by parts is learning what to choose for u and dv from the given integral. This is something you will develop a feel for as you get experience working these problems. However, there are some guidelines you can use as you are learning.
General Guidelines
1. Choose the dv term to make integration easy to get v since integration is usually more difficult than taking the derivative.
2. If you have a natural log term, always choose it to be equal to u. Remember, \((\ln x)'=1/x\) but \(\int{\ln(x)~dx} \neq 1/x \). [ one of the practice problems solves the integral \(\int{\ln(x)~dx}\) demonstrating what happens when you don't do this ]
3. If you have a polynomial, set it equal to u since taking the derivative to get du will reduce the power by one. [ the reduction formula video below shows why this works ]
Specific Guidelines
There is a rule of thumb called the LIATE rule that suggests which function to choose for u. By going down the list, try choosing the first function you come across as u. This rule does not work ALL the time but it is a good place to start as you are learning integration by parts.
LIATE Rule  

L  logarithmic functions like \(\ln(x)\) 
I  inverse trig functions like \(\arcsin(x)\) 
A  algebraic functions like polynomials 
T  trigonometric functions like \(\sin(x)\) 
E  exponential functions like \(e^{3x}\) 
source: wikipedia 
Other Situations That You Will See
 You will come across an integral where you may need to use integration by parts multiple times. This is especially true when you have a polynomial. [ the reduction formula video below demonstrates this ]
 Another case you will run into is that, after using integration by parts once or more, you may end up with the same integral you started out with. It looks like you don't accomplish anything or it loops back on itself. However, there is a way to solve these types of integrals. [ one of the practice problems solves the integral \(\int{e^{2x}\sin(x)~dx}\) showing this ]
Reduction Formulas 

Here are three videos showing how to derive reduction formulas. Although you will may never use the results of these videos directly, they are very good videos to watch to help you understand integration by parts.
In the first video, he derives the reduction formula for \( \int{ x^n e^x ~dx} \) and, in the process, demonstrates two of the main points discussed above, (1) choosing u as the polynomial reduces the power in the next step, and (2) doing integration by parts multiple times is sometimes necessary.
Dr Chris Tisdell  Reduction formula: integration  
In this video, she derives the reduction formula for \( \int{ (\ln x)^n ~dx} \).
Krista King Math  Integration by parts to prove the reduction formula  
In this video, he derives the reduction formula for \( \int{ \sec^n(x) ~dx} \).
Although the notation gets a bit messy, this is a good video demonstrating the use of integration by parts. If you get lost, just try to get the main points he is making.
PatrickJMT  Reduction Formula for: Integral of [sec(x)]^n dx  
Okay, time for you to try some practice problems.
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Practice Problems 

Instructions   Unless otherwise instructed, use integration by parts to evaluate these integrals, giving your answers in exact form.
Level A  Basic 
Practice A03  

\(\displaystyle{\int{xe^{x}~dx}}\)  
solution 
Practice A05  

\(\displaystyle{\int{x~5^x~dx}}\)  
solution 
Practice A06  

\(\displaystyle{\int_{1}^{2}{\frac{\ln(x)}{x^2}~dx}}\)  
solution 
Practice A07  

\(\displaystyle{\int{x^3\ln(x)~dx}}\)  
solution 
Practice A08  

\(\displaystyle{\int{x\cos(3x)~dx}}\)  
solution 
Practice A09  

\(\displaystyle{\int{xe^{2x}dx}}\)  
solution 
Practice A10  

\(\displaystyle{\int{\arctan(x)~dx}}\)  
solution 
Practice A11  

\(\displaystyle{\int{xe^x~dx}}\)  
solution 
Practice A12  

\(\displaystyle{\int{x^7\ln(x)~dx}}\)  
solution 
Level B  Intermediate 
Practice B03  

\(\displaystyle{\int{(\ln x)^2dx}}\)  
solution 
Practice B04  

\(\displaystyle{\int{e^{2x}\sin(x)~dx}}\)  
solution 
Practice B05  

\(\displaystyle{\int{x^2e^xdx}}\)  
solution 
Practice B06  

\(\displaystyle{\int{e^{7x}\cos(2x)~dx}}\)  
solution 
Practice B07  

\(\displaystyle{\int{x^3e^xdx}}\)  
solution 
Practice B08  

\(\displaystyle{\int{e^{2\theta}\sin(3\theta)~d\theta}}\)  
solution 
Practice B09  

\(\displaystyle{\int_{\sqrt{\pi/2}}^{\sqrt{\pi}}{\theta^3\cos(\theta^2)~d\theta}}\)  
solution 
Level C  Advanced 