Integration by Parts

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Integration by Parts is a method of integration, often useful when an integrand can be expressed as a multiplication of two functions.

The method uses the formula

$\int u{\rm d}v = uv - \int v{\rm d}u$ , where $u \ \text{and} \ v$ are the two functions.

Example

Evaluate the following integral

$\int x\text{log}x {\rm d}x$

Solution

We apply the formula $\int u{\rm d}v = uv - \int v{\rm d}u$ , where $u = \text{log}x, \ \text{and} \ {\rm d}v = x \ \text{which implies} \ v = \frac{1}{2}x^{2}.$

From $u = \text{log}x$ , we have ${\rm d}u = \frac{1}{x\text{ln}(10)}{\rm d}x$ . To find out why this is true:

$u = \text{log}x = \text{log}_{10}x$ ,

$10^{u} = x$

$u\text{ln}(10) = \text{ln}x$

Differentiating both sides,

$\text{ln}(10){\rm d}u = \frac{1}{x}{\rm d}x$ (you can make $u$ subject and then take derivative)

$\frac{{\rm d}u}{{\rm d}x} = \frac{1}{x\text{ln}(10)} \ \implies {\rm d}u = \frac{1}{x\text{ln}(10)}{\rm d}x$ .

Now,

$\int x \text{log}x {\rm d}x = \frac{1}{2}x^{2}\text{log}x - \int \frac{1}{2}x^{2}\frac{1}{x\text{ln}(10)}{\rm d}x = \frac{1}{2}x^{2}\text{log}x - \frac{1}{2\text{ln}(10)}\int x{\rm d}x$

$= \frac{1}{2}x^{2}\text{log}x - \frac{1}{2\text{ln}(10)}\left( \frac{1}{2}x^2 + C\right)$ , where $C$ is constant.

Simplifying, we get

$\int x\text{log}x {\rm d}x = \frac{1}{2}x^2 \text{log}x - \frac{1}{4\text{ln}(10)}x^2 + C$ .

Any question/comment?

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