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}$${\cos ^n}x)dx$
11. Integral of type $\int {\frac{{{x^2} \pm 1}}{{{x^4} + k{x^2} + 1}}dx} $
12. Integration of irrational algebraic functions
13.0 Integral of type $\int {{x^m}{{\left( {a + b{x^n}} \right)}^p}dx} $
14.0 Reduction formulae
8.2 Type B: Linear and repeating
8.2 Type B: Linear and repeating
8.3 Type C: Quadratic and non-repeating
8.4 Type D: Quadratic and repeating
Let us assume, $$g(x) = {(x - a)^k}(x - {a_1})(x - {a_2})...(x - {a_r})$$ Therefore, rational algebraic function is written as, $$\frac{{f(x)}}{{g(x)}} = \frac{{{A_1}}}{{x - a}} + \frac{{{A_2}}}{{{{(x - a)}^2}}} + ... + \frac{{{A_k}}}{{{{(x - a)}^k}}} + \frac{{{B_1}}}{{x - {a_1}}} + \frac{{{B_2}}}{{x - {a_2}}} + ... + \frac{{{B_r}}}{{x - {a_r}}}$$
Question 7. Evaluate $$\int {\frac{{x - 5}}{{{{(x - 2)}^2}}}} dx$$
Solution: From the given integral, $f(x)=x+5$ and $g(x) = {(x - 2)^2}$. Therefore, $$\frac{{f(x)}}{{g(x)}} = \frac{{x + 5}}{{{{(x - 2)}^2}}}$$
Let,$$\begin{equation} \begin{aligned} \frac{{x + 5}}{{{{(x - 2)}^2}}} = \frac{A}{{x - 2}} + \frac{B}{{{{(x - 2)}^2}}} \\ \Rightarrow x + 5 = A(x - 2) + B \\\end{aligned} \end{equation} $$
On comparing, we get $A=1$ and $B=7$. Therefore, $$\begin{equation} \begin{aligned} I = \int {\frac{{x + 5}}{{{{(x - 2)}^2}}}} dx \\ I = \int {\frac{1}{{x - 2}}} dx + \int {\frac{7}{{{{(x - 2)}^2}}}} dx \\ I = \ln (x - 2) - \frac{7}{{x - 2}} + C \\\end{aligned} \end{equation} $$
