Nguyên hàm - Định nghĩa và tính chất

\(\int {\left( {{x^2} + 4} \right)dx}  = \dfrac{{{x^3}}}{3} + 4x + C\).

Ta có \(\int {\left( {{e^x} + 2} \right)dx}  = {e^x} + 2x + C\).

Ta có \(f\left( x \right) = \dfrac{{{x^2} - 2x + 1}}{{x - 2}} = x + \dfrac{1}{{x - 2}}\).

\( \Rightarrow \int {f\left( x \right)dx}  = \int {\left( {x + \dfrac{1}{{x - 2}}} \right)dx} \)$=\int xdx + \int { \dfrac{1}{{x - 2}}dx  }  $\( = \dfrac{{{x^2}}}{2} + \ln \left| {x - 2} \right| + C\).

Xét hàm số: \(f\left( x \right) = \dfrac{1}{{{x^2} - 2x}}\) trong \(\left( {2; + \infty } \right)\) ta có:

\( \dfrac{1}{{{x^2} - 2x}}=\dfrac{1}{{x\left( {x - 2} \right)}}=\dfrac{1}{{2\left( {x - 2} \right)}} - \dfrac{1}{{2x}} \)

Khi đó:

\(\begin{array}{l}\,\,\,\,\int {f\left( x \right)dx}  = \int {\dfrac{1}{{{x^2} - 2x}}dx} \\ = \int {\left( {\dfrac{1}{{2\left( {x - 2} \right)}} - \dfrac{1}{{2x}}} \right)dx} \\ = \dfrac{1}{2}\ln \left| {x - 2} \right| - \dfrac{1}{2}\ln \left| x \right| + C\\ = \dfrac{1}{2}\left( {\ln \left| {x - 2} \right| - \ln \left| x \right|} \right) + C.\end{array}\)

Vì \(x \in \left( {2; + \infty } \right) \Rightarrow \left\{ \begin{array}{l}\left| {x - 2} \right| = x - 2\\\left| x \right| = x\end{array} \right.\)

Do đó \(\int {f\left( x \right)dx}  = \dfrac{{\ln \left( {x - 2} \right) - \ln x}}{2} + C\).