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Example Evaluate the following integral．

$\displaystyle \int \frac{2x + 1}{x^2 + x} dx$

Answer 関数の積分を求めるうえで重要なことは，つぎのことである．

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． $\int e^{x}dx = e^{x} + c, \int \sin{x} = -\cos{x}+c, \int \frac{1}{x}dx = \log\vert x\vert + c$



Given the integrand $\frac{2x + 1}{x^2 + x}$, we will tranform this into something easy to integrate．To do so，let $t = x^2 + x$. Then $dt = (2x+1)dx$ and $\int \frac{1}{t}dt$ which is manageable．


    $\displaystyle \int \frac{2x + 1}{x^2 + x} dx　\ \left(\begin{array}{ll} t = x^2 + x\\ dt = (2x+1)dx \end{array}\right.$  
  $\displaystyle =$ $\displaystyle \int \frac{1}{t}dt = \log\vert t\vert + c$  
  $\displaystyle =$ $\displaystyle \log\vert x^2 + x\vert + c$