Tích phân hàm hữu tỉ

Loại 1: Tính $I =\int\limits_\alpha ^\beta  {\dfrac{{dx}}{{{{\left( {ax + b} \right)}^k}}}}  = \dfrac{1}{a}\int\limits_\alpha ^\beta  {{{\left( {ax + b} \right)}^{ - k}}.adx}  = \dfrac{1}{{a\left( {1 - k} \right)}}.{\left( {ax + b} \right)^{ - k + 1}}\left| {_\alpha ^\beta } \right.$

Loại 2: Tính $I=\int\limits_\alpha ^\beta  {\dfrac{{dx}}{{a{x^2} + bx + c}}\left( {a \ne 0} \right)}$

+Nếu $\Delta>0$

$\dfrac{1}{a{{x}^{2}}+bx+c}=\dfrac{1}{a(x-{{x}_{1}})(x-{{x}_{2}})}=\dfrac{1}{a({{x}_{1}}-{{x}_{2}})}\left(\dfrac{1}{x-{{x}_{1}}}-\dfrac{1}{x-{{x}_{2}}} \right)$

thì $I=\dfrac{1}{a({{x}_{1}}-{{x}_{2}})}\int\limits_{\alpha }^{\beta }{\left( \dfrac{1}{x-{{x}_{1}}}-\dfrac{1}{x-{{x}_{2}}}\right)}dx=\dfrac{1}{a({{x}_{1}}-{{x}_{2}})}\left[ \ln \left| x-{{x}_{1}} \right|-\ln \left| x-{{x}_{2}} \right| \right]\left| _{\alpha }^{\beta}= \right.\dfrac{1}{a({{x}_{1}}-{{x}_{2}})}\ln \left| \dfrac{x-{{x}_{1}}}{x-{{x}_{2}}} \right|\left| _{\alpha }^{\beta } \right.$

+Nếu $\Delta=0:$

$\dfrac{1}{a{{x}^{2}}+bx+c}=\dfrac{1}{a{{(x-{{x}_{0}})}^{2}}}\quad \left( {{x}_{0}}=\dfrac{-b}{2a} \right)$

thìI=$\int\limits_{\alpha }^{\beta }{\dfrac{dx}{a{{x}^{2}}+bx+c}=}\dfrac{1}{a}\int\limits_{\alpha }^{\beta }{\dfrac{dx}{{{(x-{{x}_{0}})}^{2}}}=-\dfrac{1}{a(x-{{x}_{0}})}\left| _{\alpha }^{\beta } \right.}$

+Nếu $\Delta <0:$

$I=\dfrac{1}{a}\int\limits_{\alpha }^{\beta }{\dfrac{dx}{{{(x+m)}^{2}}+{{n}^{2}}}}$;ta đổi biến số:$x+m=n\tan t$