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Example Evaluate the following integral．

$\displaystyle \int \log{x} dx$

Answer To find the integral, it is important to know about the integration by parts.．

1. The integration of a sum is the sum of the integration． $\int (f(x) \pm g(x))dx = \int f(x)dx \pm \int g(x)dx$
2. The integration of a constant multiple is the constant times the integral． $\int (\alpha f(x))dx = \alpha \int f(x)$
3. Integration By Substitution． $\int f(x)dx = \int f(\Phi(t))(\Phi(t))'dt $
4. Integration By Parts. $\int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx$
5. Partial Fraction． $\frac{3x+2}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}$
6. Basic integration formula． $\int e^{x}dx = e^{x} + c, \int \sin{x} = -\cos{x}+c, \int \frac{1}{x}dx = \log\vert x\vert + c$



\begin{displaymath}\begin{array}{ll} {\rm Put} u = \log{x}, & dv = dx\\ du = \frac{1}{x}dx, & v = \int dv = \int dx = x + c \end{array} \end{displaymath}

Then
$\displaystyle \int \underbrace{\log{x}}_{u} \underbrace{dx}_{dv}$ $\displaystyle =$ $\displaystyle \underbrace{(x + c)}_{v} \underbrace{\log{x}}_{u} - \int \underbrace{(x + c)}_{v} \underbrace{\frac{1}{x} dx}_{du}$  
  $\displaystyle =$ $\displaystyle x \log{x} + c\log{x} - x - c\log{x} + c$  
  $\displaystyle =$ $\displaystyle x \log{x} - x + c$  

Thus to evaluate $v = \int dv$, you can ignore $c$.