TÍCH PHÂN HÀM PHÂN THỨC

\[\int \frac{{f\left( x \right)}}{{g\left( x \right)}}dx\]

+ Tích phân cơ bản\(\displaystyle {\rm{\;I}} = \mathop \int \limits_{\rm{a}}^{\rm{b}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}}{\rm{dx}}\) , trong đó \(\displaystyle {\rm{f}}\left( {\rm{x}} \right);{\rm{g}}\left( {\rm{x}} \right)\) là các đa thức:

+ Cách giải:

- Nếu bậc của f(x) lớn hơn hay bằng g(x) thì chia f(x) cho g(x) để tách thương (phần nguyên), nên có thể coi bậc f(x) nhỏ hơn bậc g(x).

- Phân tích ${\text{g}}\left( {\text{x}} \right)$ thành nhân tử gồm những nhân tử là các nhị thức bậc nhất và tam thức bậc hai vô nghiệm:

\(\displaystyle{\rm{g}}\left( {\rm{x}} \right) = {\left( {{{\rm{a}}_1}{\rm{x}} + {{\rm{b}}_1}} \right)^{{{\rm{n}}_1}}}.{\left( {{{\rm{a}}_2}{\rm{x}} + {{\rm{b}}_2}} \right)^{{{\rm{n}}_2}}} \ldots\)\(\displaystyle {\left( {{{\rm{\alpha }}_1}{{\rm{x}}^2} + {{\rm{\beta }}_1}{\rm{x}} + {{\rm{\gamma }}_1}} \right)^{{{\rm{m}}_1}}}.{\left( {{{\rm{\alpha }}_2}{{\rm{x}}^2} + {{\rm{\beta }}_2}{\rm{x}} + {{\rm{\gamma }}_2}} \right)^{{{\rm{m}}_2}}} \ldots \)

Dùng hệ số bất định ta tách được thành:

\(\displaystyle \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{{{{\rm{A}}_{11}}}}{{{{\left( {{{\rm{a}}_1}{\rm{x}} + {{\rm{b}}_1}} \right)}^{{{\rm{n}}_1}}}}}\)\(\displaystyle + \frac{{{{\rm{A}}_{12}}}}{{{{\left( {{{\rm{a}}_1}{\rm{x}} + {{\rm{b}}_1}} \right)}^{{{\rm{n}}_1} - 1}}}} + \ldots \)\(\displaystyle + \frac{{{{\rm{A}}_{1{{\rm{n}}_1}}}}}{{{{\rm{a}}_1}{\rm{x}} + {{\rm{b}}_1}}} + \frac{{{{\rm{A}}_{21}}}}{{{{\left( {{{\rm{a}}_2}{\rm{x}} + {{\rm{b}}_2}} \right)}^{{{\rm{n}}_2}}}}} + \frac{{{{\rm{A}}_{22}}}}{{{{\left( {{{\rm{a}}_2}{\rm{x}} + {{\rm{b}}_2}} \right)}^{{{\rm{n}}_2} - 1}}}}\)\(\displaystyle + \ldots + \frac{{{{\rm{A}}_{2{{\rm{n}}_2}}}}}{{{{\rm{a}}_2}{\rm{x}} + {{\rm{b}}_2}}} + \ldots + \ldots\)

Tách thành tổng các tích phân rồi tính từng tích phân một.

Ví dụ 1: Tính tích phân:

a)\(\displaystyle {\rm{I}} = \int \frac{{3{\rm{x}} + 2}}{{{{\rm{x}}^2} - 1}}{\rm{dx}}\)

b)\(\displaystyle {\rm{I}} = \mathop \int \limits_0^1 \frac{{2{\rm{x}} + 1}}{{{{\rm{x}}^4} + {{\rm{x}}^2} + 1}}{\rm{dx}}\)

c)\(\displaystyle {\rm{I}} = \mathop \int \limits_0^1 \frac{{\left( {{{\rm{x}}^2} + {\rm{x}} + 2} \right)}}{{{{\left( {{\rm{x}} + 1} \right)}^2}\left( {{{\rm{x}}^2} + {\rm{x}} + 1} \right)}}{\rm{dx\;}}\)

Giải:

a) Ta có: \(\displaystyle \frac{{3{\rm{x}} + 2}}{{{{\rm{x}}^2} - 1}} = \frac{{3{\rm{x}} + 2}}{{\left( {{\rm{x}} - 1} \right)\left( {{\rm{x}} + 1} \right)}} = \frac{{\rm{a}}}{{{\rm{x}} - 1}} + \frac{{\rm{b}}}{{{\rm{x}} + 1}}\) . Ta cần tìm ra ${\text{a}}$ và ${\text{b}}$ (chắc chắn tìm được) bằng phương pháp cân bằng hệ số.

Ta có: \(\displaystyle \frac{{\rm{a}}}{{{\rm{x}} - 1}} + \frac{{\rm{b}}}{{{\rm{x}} + 1}} = \frac{{\left( {{\rm{a}} + {\rm{b}}} \right){\rm{x}} + {\rm{a}} - {\rm{b}}}}{{\left( {{\rm{x}} - 1} \right)\left( {{\rm{x}} + 1} \right)}}\) . Vậy \(\displaystyle \frac{{3{\rm{x}} + 2}}{{\left( {{\rm{x}} - 1} \right)\left( {{\rm{x}} + 1} \right)}} = \frac{{\rm{a}}}{{{\rm{x}} - 1}} + \frac{{\rm{b}}}{{{\rm{x}} + 1}}\)\(\displaystyle \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{a + b = 3}\\{a - b = 2}\end{array}{\rm{\;\;}} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{a = \frac{5}{2}}\\{b = \frac{1}{2}}\end{array}} \right.{\rm{\;}}} \right.\) .

Vậy ta có \(\displaystyle {\rm{I}} = \int \left( {\frac{5}{2}.\frac{1}{{{\rm{x}} - 1}} + \frac{1}{2}.\frac{1}{{{\rm{x}} + 1}}} \right){\rm{dx}}\)\(\displaystyle = \frac{5}{2}\ln \left| {{\rm{x}} - 1} \right| + \frac{1}{2}\ln \left| {{\rm{x}} + 1} \right| + {\rm{C}}\).

b) Ta có \(\displaystyle \frac{{2{\rm{x}} + 1}}{{{{\rm{x}}^4} + {{\rm{x}}^2} + 1}} = \frac{{2{\rm{x}} + 1}}{{\left( {{{\rm{x}}^2} - {\rm{x}} + 1} \right)\left( {{{\rm{x}}^2} + {\rm{x}} + 1} \right)}}\) \(\displaystyle = \frac{{{\rm{ax}} + {\rm{b}}}}{{{{\rm{x}}^2} - {\rm{x}} + 1}} + \frac{{{\rm{cx}} + {\rm{d}}}}{{{{\rm{x}}^2} + {\rm{x}} + 1}}\) .

Ta tìm ra \(\displaystyle {\rm{a}};{\rm{b}};{\rm{c}};{\rm{d}}\) bằng phương pháp cân bằng hệ số.

Ta có: \(\displaystyle \frac{{{\rm{ax}} + {\rm{b}}}}{{{{\rm{x}}^2} - {\rm{x}} + 1}} + \frac{{{\rm{cx}} + {\rm{d}}}}{{{{\rm{x}}^2} + {\rm{x}} + 1}}\)\(\displaystyle = \frac{{\left( {{\rm{a}} + {\rm{c}}} \right){{\rm{x}}^3} + \left( {{\rm{a}} + {\rm{b}} + {\rm{d}} - {\rm{c}}} \right){{\rm{x}}^2} + \left( {{\rm{a}} + {\rm{b}} + {\rm{c}} - {\rm{d}}} \right){\rm{x}} + \left( {{\rm{b}} + {\rm{d}}} \right)}}{{\left( {{{\rm{x}}^2} - {\rm{x}} + 1} \right)\left( {{{\rm{x}}^2} + {\rm{x}} + 1} \right)}}\) .
Vậy hệ số cân bằng đẳng thức ban đầu khi và chỉ khi \(\displaystyle \left\{ {\begin{array}{*{20}{c}}{a + c = 0}\\{a + b + d - c = 0}\\{a + b + c - d = 2}\\{b + d = 1}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{a = - \frac{1}{2}}\\{b = \frac{3}{2}}\\{c = \frac{1}{2}}\\{d = - \frac{1}{2}}\end{array}} \right.\)

Vậy \(\displaystyle \frac{{2{\rm{x}} + 1}}{{{{\rm{x}}^4} + {{\rm{x}}^2} + 1}} = \frac{1}{2}\frac{{ - {\rm{x}} + 3}}{{{{\rm{x}}^2} - {\rm{x}} + 1}} + \frac{1}{2}.\frac{{{\rm{x}} - 1}}{{{{\rm{x}}^2} + {\rm{x}} + 1}}\) . Do đó, \(\displaystyle {\rm{I}} = \frac{1}{2}\mathop \int \limits_0^1 \frac{{ - {\rm{x}} + 3}}{{{{\rm{x}}^2} - {\rm{x}} + 1}}{\rm{dx}} + \frac{1}{2}\mathop \int \limits_0^1 \frac{{{\rm{x}} - 1}}{{{{\rm{x}}^2} + {\rm{x}} + 1}}{\rm{dx}}\)

c) Ta \(\displaystyle \frac{{\left( {{{\rm{x}}^2} + {\rm{x}} + 2} \right)}}{{{{\left( {{\rm{x}} + 1} \right)}^2}\left( {{{\rm{x}}^2} + {\rm{x}} + 1} \right)}}\)\(\displaystyle = \frac{{\rm{A}}}{{{{\left( {{\rm{x}} + 1} \right)}^2}}} + \frac{{\rm{B}}}{{{\rm{x}} + 1}} + \frac{{{\rm{ax}} + {\rm{b}}}}{{{{\rm{x}}^2} + {\rm{x}} + 1}}\).

Bằng cách như trên ta tìm ra được \(\displaystyle {\rm{A}};{\rm{B}};{\rm{a}};{\rm{b}}\) .

* Bây giờ ta tính tích phân \(\displaystyle \mathop \int \limits_{\rm{u}}^{\rm{v}} \frac{{{\rm{\alpha x}} + {\rm{\beta }}}}{{{\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}}}}{\rm{dx}}\) với \(\displaystyle {\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}}\) là tam thức bậc 2 vô nghiệm.

Ta có: \(\displaystyle \frac{{{\rm{\alpha x}} + {\rm{\beta }}}}{{{\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}}}} = \frac{{\frac{{\rm{\alpha }}}{{2{\rm{a}}}}\left( {2{\rm{ax}} + {\rm{b}}} \right) + {\rm{\beta }} - \frac{{\rm{\alpha }}}{2}.\frac{{\rm{b}}}{{\rm{a}}}}}{{{\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}}}}\)\(\displaystyle = \frac{{\rm{\alpha }}}{{2{\rm{a}}}}.\frac{{{{\left( {{\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}}} \right)}^{\rm{'}}}}}{{{\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}}}} + \frac{{\rm{C}}}{{{\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}}}}\)

= \(\displaystyle \frac{{\rm{\alpha }}}{{2{\rm{a}}}}.\frac{{{{\left( {{\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}}} \right)}^{\rm{'}}}}}{{{\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}}}} + \frac{{\rm{C}}}{{\rm{a}}}.\frac{1}{{{{\left( {{\rm{x}} + \frac{{\rm{b}}}{{2{\rm{a}}}}} \right)}^2} + {{\left( {\sqrt { - \frac{{\rm{\Delta }}}{{2{\rm{a}}}}} } \right)}^2}}}\) , Với \(\displaystyle {\rm{C}} = {\rm{\beta }} - \frac{{\rm{\alpha }}}{2}.\frac{{\rm{b}}}{{\rm{a}}}\).

Vậy \(\displaystyle \mathop \int \limits_{\rm{u}}^{\rm{v}} \frac{{{\rm{\alpha x}} + {\rm{\beta }}}}{{{\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}}}}{\rm{dx}} = \frac{{\rm{\alpha }}}{{2{\rm{a}}}}\mathop \int \limits_{\rm{u}}^{\rm{v}} \frac{{{{\left( {{\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}}} \right)}^{\rm{'}}}}}{{{\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}}}}{\rm{dx}}\)\(\displaystyle + \frac{{\rm{C}}}{{\rm{a}}}\mathop \int \limits_{\rm{u}}^{\rm{v}} \frac{{{\rm{dx}}}}{{{{\left( {{\rm{x}} + \frac{{\rm{b}}}{{2{\rm{a}}}}} \right)}^2} + {{\left( {\sqrt { - \frac{{\rm{\Delta }}}{{2{\rm{a}}}}} } \right)}^2}}}\) \(\displaystyle = {\rm{ln}}\left| {{\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}}} \right| + \frac{{\rm{C}}}{{\rm{a}}}\mathop \int \limits_{\rm{u}}^{\rm{v}} \frac{{{\rm{dx}}}}{{{{\left( {{\rm{x}} + \frac{{\rm{b}}}{{2{\rm{a}}}}} \right)}^2} + {{\left( {\sqrt { - \frac{{\rm{\Delta }}}{{2{\rm{a}}}}} } \right)}^2}}}\). Tích phân còn lại \(\displaystyle \frac{C}{a}\mathop \int \limits_u^v \frac{{dx}}{{{{\left( {x + \frac{b}{{2a}}} \right)}^2} + {{\left( {\sqrt { - \frac{\Delta }{{2a}}} } \right)}^2}}}\) ta đặt X = \(\displaystyle x + \frac{b}{{2a}}{\rm{\;}};A = \sqrt { - \frac{\Delta }{{2a}}} \) ta đưa về tích phân \(\displaystyle \frac{C}{a}\mathop \int \limits_{u{\rm{'}}}^{v{\rm{'}}} \frac{{dX}}{{{X^2} + {A^2}}}\). Lại đặt \(\displaystyle X = A\tan t\) theo phương pháp của Bài 2 dưới đây.

Ví dụ 2: Tính tích phân \(\displaystyle I = \mathop \int \limits_0^1 \frac{{x - 1}}{{{x^2} + x + 1}}dx\).

Giải: \(\displaystyle = \mathop \int \limits_0^1 \frac{{\frac{1}{2}\left( {2x + 1} \right)- \frac{3}{2}}}{{{x^2} + x + 1}}dx \)\(\displaystyle= \frac{1}{2}\mathop \int \limits_0^1 \frac{{d\left( {{x^2} + x + 1} \right)}}{{{x^2} + x + 1}}\)\(\displaystyle - \frac{3}{2}\mathop \int \limits_0^1 \frac{{dx}}{{{{\left( {x + \frac{1}{2}} \right)}^2} + \frac{3}{4}}} \)\(\displaystyle = \frac{1}{2}\ln \left| {{x^2} + x + 1} \right|\left. \right|_0^1 - \frac{3}{2}\mathop \int \limits_0^1 \frac{{dx}}{{{{\left( {x + \frac{1}{2}} \right)}^2} + \frac{3}{4}}}\) . Đặt \(\displaystyle x + \frac{1}{2} = \frac{{\sqrt 3 }}{2}\tan t \Rightarrow \left\{ {\begin{array}{*{20}{c}}{x = 0 \Rightarrow t = \frac{\pi }{6}}\\{x = 1 \Rightarrow t = \frac{\pi }{3}}\\{dx = \frac{{\sqrt 3 }}{2}\left( {1 + {{\tan }^2}t} \right)dt}\end{array}} \right.\)

Vậy \(\displaystyle \mathop \int \limits_0^1 \frac{{dx}}{{{{\left( {x + \frac{1}{2}} \right)}^2} + \frac{3}{4}}} = \mathop \int \limits_{\pi /6}^{\pi /3} \frac{{\frac{{\sqrt 3 }}{2}\left( {1 + {{\tan }^2}t} \right)dt}}{{\frac{3}{4}\left( {1 + {{\tan }^2}t} \right)}}\)\(\displaystyle = \frac{{2\sqrt 3 }}{3}\mathop \int \limits_{\pi /6}^{\pi /3} dt = \frac{{2\sqrt 3 }}{3}.\frac{\pi }{6} = \frac{{\sqrt 3 \pi }}{9}\)\(\displaystyle \to = \frac{1}{2}.\ln 3 - \frac{{\sqrt 3 \pi }}{6}\) .