## Integration by Parts

Integration by parts is a method that allows us to integrate the product of two function such as

∫2xe^{3x}dx

where 2x is one function and e^{3x} is another

To do this we use the formula

**∫(u d/dx) dx = uv – ∫(v du/dx) dx**

where u and v are both functions of x, 2x and e^{3x} in the above example.

**Proof of Integration by Parts Formula**

This can be shown by considering the product rule for differentiation as shown below

the product rule states that

d(uv)/dx = u dv/dx + v du/dx

If we now integrate both side we get

uv = ∫( u dv/dx) dx + ∫(v du/dx) dx

since integration is the opposite of differentiation

now we can simply rearrange this to get

∫(u d/dx) dx = uv – ∫(v du/dx) dx

**Simple Example (without limits)**

To demonstrate this formula we shall integrate the example above

∫2xe^{3x}dx

To do this we first need to pick which function (2x or e^{3x}) to allocate to u and which to dv/dx. Picking the correct function is crucial to integration by parts since the method works by allowing us the differentiate the part of the function which is difficult to integrate. In order for this to work it is useful to use a function that will become simpler in some way. For example differentiating e^{3x} gives 3e^{3x} so making the product no easier to integrate, however integrating 2x gives 2 which does make it easier to integrate since we can take the 2 outside as a factor.

So we will let u=2x and dv/dx=e^{3x}

we then differentiate 2x to get 2

and integrate e^{3x} to e^{3x}/3

(it can often be useful to put these in a table with columns u,v and the function on the first row and derivative on the second row)

These can then be put into the formula to find the integral

∫2xe^{3x}dx = 2xe^{3x}/3 – ∫e^{3x}2/3 dx

∫2xe^{3x}dx = 2xe^{3x}/3 – 2e^{3x}/9 + c

**Example with Limits**

usually, however integration is required to be carried out between limits. To perform integration by parts between limits is quite simple. You takes the value of uv as usual between limits and then set the limits of ∫v du/dx dx to the limits of the original integral. To demonstrate this we will integrate the above example between 0 and 1

so

∫^{1}_{0} 2x e^{3x} dx = [2xe^{3x}/3]^{1}_{0} – ∫^{1}_{0} e^{3x}2/3 dx

=2e^{3} – 0 – 2e^{3}/9 + 2/9 = (16e^{3} + 2) / 9

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