Maths > Indefinite Integrals > 8.0 Integration using partial fractions
Indefinite Integrals
1.0 Introduction
2.0 Methods of Integration
3.0 Integration by parts
4.0 Integral of the type $\int {{e^x}\left\{ {f(x) + f'(x)} \right\}dx} $
5.0 Integral of the type $\int {\frac{{dx}}{{a{x^2} + bx + c}},\int {\frac{{dx}}{{\sqrt {a{x^2} + bx + c} }},\int {\sqrt {a{x^2} + bx + c} } dx} } $
6.0 Integral of the type $\int {\frac{{px + q}}{{a{x^2} + bx + c}}dx,} \int {\frac{{px + q}}{{\sqrt {a{x^2} + bx + c} }}dx,} \int {(px + q)} \sqrt {a{x^2} + bx + c} dx$
7.0 Integral of the type $\int {\frac{{a{x^2} + bx + c}}{{(p{x^2} + qx + r)}}dx,\int {\frac{{a{x^2} + bx + c}}{{\sqrt {p{x^2} + qx + r} }}dx} } ,\int {(a{x^2} + bx + c)} \sqrt {p{x^2} + qx + r} dx$
8.0 Integration using partial fractions
8.1 Type A: Linear and non-repeating
8.2 Type B: Linear and repeating
8.3 Type C: Quadratic and non-repeating
8.4 Type D: Quadratic and repeating
9.0 Integration of trigonometric functions
10.0 Integral of type $\int {({{\sin }^m}x}$${\cos ^n}x)dx$
11. Integral of type $\int {\frac{{{x^2} \pm 1}}{{{x^4} + k{x^2} + 1}}dx} $
12. Integration of irrational algebraic functions
13.0 Integral of type $\int {{x^m}{{\left( {a + b{x^n}} \right)}^p}dx} $
14.0 Reduction formulae
8.1 Type A: Linear and non-repeating
8.2 Type B: Linear and repeating
8.3 Type C: Quadratic and non-repeating
8.4 Type D: Quadratic and repeating
Let us assume $$g(x) = (x - {a_1})(x - {a_2})...(x - {a_n})$$
Therefore, rational algebraic function is written as, $$\frac{{f(x)}}{{g(x)}} = \frac{{{A_1}}}{{x - {a_1}}} + \frac{{{A_2}}}{{x - {a_2}}} + \frac{{{A_3}}}{{x - {a_3}}} + ... + \frac{{{A_n}}}{{x - {a_n}}}$$ where ${A_1}$, ${A_2}$, ${A_3}$,...,${A_n}$ are constants and can be calculated by comparing the numerators on both the sides which is explained with the help of example.
Question 6. Evaluate $$\int {\frac{{3x + 2}}{{{x^3} - 6{x^2} + 11x - 6}}dx} $$
Solution: From the given integral, $f(x) = 3x + 2$ and $g(x) = {x^3} - 6{x^2} + 11x - 6$. Therefore, $$\frac{{f(x)}}{{g(x)}} = \frac{{3x + 2}}{{{x^3} - 6{x^2} + 11x - 6}} = \frac{{3x + 2}}{{(x - 1)(x - 2)(x - 3)}}$$
Let,$$\Rightarrow \frac{{3x + 2}}{{(x - 1)(x - 2)(x - 3)}} = \frac{A}{{x - 1}} + \frac{B}{{x - 2}} + \frac{C}{{x - 3}} $$ $$ \Rightarrow \frac{{3x + 2}}{{(x - 1)(x - 2)(x - 3)}} = \frac{{A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)}}{{(x - 1)(x - 2)(x - 3)}} $$ $$ \Rightarrow 3x + 2 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)\quad ...(1) $$
Put $x-1=0$ i.e., $x=1$ we get,
$$5=A(1-2)(1-3)$$$$ \Rightarrow A = \frac{5}{2}$$
Similarly, by putting $x-2=0$ and $x-3=0$, we get $B=-8$ and $C = \frac{{11}}{2}$
$$\frac{{3x + 2}}{{(x - 1)(x - 2)(x - 3)}} = \frac{5}{{x - 1}} - \frac{8}{{x - 2}} + \frac{{11}}{{2(x - 3)}}$$
Now,
$$\begin{equation} \begin{aligned} I = \int {\frac{{3x + 2}}{{(x - 1)(x - 2)(x - 3)}}} dx \\ I = \int {\frac{5}{{x - 1}}dx} - \int {\frac{8}{{x - 2}}dx} + \int {\frac{{11}}{{2(x - 3)}}dx} \\ I = 5\ln (x - 1) - 8\ln (x - 2) + \frac{{11}}{2}\ln (x - 3) + C \\\end{aligned} \end{equation} $$
