Tính nguyên hàm x^2 − 1 / ( x^2 + 1)^2 d x ?

Ta có:\[\frac{{{x^2} - 1}}{{{{\left( {{x^2} + 1} \right)}^2}}} = \frac{{2{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}} - \frac{1}{{{x^2} + 1}}\]

\[ \Rightarrow \smallint \frac{{{x^2} - 1}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx = \smallint \frac{{2{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx - \smallint \frac{1}{{{x^2} + 1}}dx\,\,\left( 1 \right)\]

Ta tính\[\smallint \frac{{2{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx = \smallint \frac{{xd\left( {{x^2} + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}}\] bằng phương pháp tích phân từng phân như sau:

Đặt\(\left\{ {\begin{array}{*{20}{c}}{u = x}\\{dv = \frac{{d({x^2} + 1)}}{{{{({x^2} + 1)}^2}}}}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = dx}\\{v = - \frac{1}{{{x^2} + 1}}}\end{array}} \right.\)

\[ \Rightarrow \smallint \frac{{xd\left( {{x^2} + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}} = - \frac{x}{{{x^2} + 1}} + \smallint \frac{{dx}}{{{x^2} + 1}} + C\,\,\left( 2 \right)\]

Từ (1) và (2) suy ra 

\[\smallint \frac{{{x^2} - 1}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx = - \frac{x}{{{x^2} + 1}} + \smallint \frac{{dx}}{{{x^2} + 1}} + C - \smallint \frac{1}{{{x^2} + 1}}dx = - \frac{x}{{{x^2} + 1}} + C.\]