Integration by parts.

    

The so-called integration by parts formula is a direct consequence of Leibniz's formula for derivation of a product (v.s. 5):

Proposition 8.2.6   \fbox{
$\int u(x) v'(x) \; dx = u(x) v(x) - \int u'(x) v(x) \; dx$
}

In each example, we have to decide who is $ u$ and who is $ v$ .

Example 8.2.7   Let $ F(x)=\int x \sin x dx$ .

We choose: \begin{displaymath}\begin{cases}u(x)=x  v(x)=-\cos x \end{cases} \Longrightarrow
\begin{cases}u'(x)=1  v'(x)=\sin x \end{cases}.\end{displaymath} Therefore: $ F(x)= - x \cos x - \int 1 \cdot (-\cos x) \; dx = - x \cos x + \int \cos x \; dx = - x \cos x + \sin x +C$ .

Example 8.2.8   Let $ F(x)=\int x \ln x \; dx$ .

In this case it is wise to choose $ u(x)= \ln x$ , so the logarithm will ``disappear'':

\begin{displaymath}\begin{cases}u(x)=\ln x  v(x)= \frac 12 x^2 \end{cases} \Longrightarrow \begin{cases}u'(x)= \frac 1x  v'(x)=x \end{cases}.\end{displaymath}    

Therefore:

$\displaystyle F(x)= \frac 12 x^2 \ln x - \int \frac 12 x^2 \cdot \frac 1x \; dx...
...rac 12 x^2 \ln x - \frac 12 \int x \; dx =\frac 12 x^2 \ln x - \frac 14 x^2 +C.$    

Thew following example uses the same technique, and its result should be remembered.

Example 8.2.9   Let $ F(x)=\int \ln x \; dx$ . Once again, we choose $ u(x)= \ln x$ :

$\displaystyle F(x)=\int \ln x \; dx = \int 1 \cdot \ln x \; dx = x \; \ln x - \int x \cdot \frac 1x \; dx = x \; \ln x - \int 1 \; dx = x \; ln x -1 +C.$    

Noah Dana-Picard 2007-12-28