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}$${\cos ^n}x)dx$
11. Integral of type $\int {\frac{{{x^2} \pm 1}}{{{x^4} + k{x^2} + 1}}dx} $
12. Integration of irrational algebraic functions
13.0 Integral of type $\int {{x^m}{{\left( {a + b{x^n}} \right)}^p}dx} $
14.0 Reduction formulae
8.3 Type C: Quadratic and non-repeating
8.2 Type B: Linear and repeating
8.3 Type C: Quadratic and non-repeating
8.4 Type D: Quadratic and repeating
Let us assume,$$g(x) = ({a_1}{x^2} + {b_1}x + {c_1})({a_2}{x^2} + {b_2}x + {c_2})...({a_r}{x^2} + {b_r}x + {c_r})(x - {d_1})(x - {d_2})...(x - {d_k})$$
Therefore, rational algebraic function is written as $$\frac{{f(x)}}{{g(x)}} = \frac{{{A_1}x + {B_1}}}{{{a_1}{x^2} + {b_1}x + {c_1}}} + \frac{{{A_2}x + {B_2}}}{{{a_2}{x^2} + {b_2}x + {c_2}}} + ... + \frac{{{A_r}x + {B_r}}}{{{a_r}{x^2} + {b_r}x + {c_r}}} + \frac{{{D_1}}}{{x - {d_1}}} + \frac{{{D_2}}}{{x - {d_2}}} + ... + \frac{{{D_k}}}{{x - {d_k}}}$$
Question 8. Evaluate $$\int {\frac{{2x - 1}}{{(x + 1)({x^2} + 2)}}dx} $$
Solution: Let us assume,
$$\begin{equation} \begin{aligned} \frac{{2x - 1}}{{(x + 1)({x^2} + 2)}} = \frac{A}{{x - 1}} + \frac{{Bx + C}}{{{x^2} + 2}} \\ \Rightarrow 2x - 1 = A({x^2} + 2) + (Bx + C)(x + 1) \\\end{aligned} \end{equation} $$
Putting $x+1=0$ i.e., $x=-1$, we get $A=-1$. Similarly, we get $B=1$ and $C=1$.
Therefore, $$\begin{equation} \begin{aligned} I = \int {\frac{{2x - 1}}{{(x + 1)({x^2} + 2)}}} dx \\ I = \int { - \frac{1}{{x - 1}}} dx + \int {\frac{{x + 1}}{{{x^2} + 2}}dx} \\ I = - \ln \left| {x - 1} \right| + \int {\frac{x}{{{x^2} + 2}}dx} + \int {\frac{1}{{{x^2} + 2}}dx} \\ I = - \ln \left| {x - 1} \right| + \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\frac{x}{{\sqrt 2 }} + \int {\frac{x}{{{x^2} + 2}}dx} \\\end{aligned} \end{equation} $$ For the third integral,
Put, $${x^2} + 2 = t \Rightarrow 2xdx = dt$$ Therefore, $$\begin{equation} \begin{aligned} I = - \ln \left| {x - 1} \right| + \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\frac{x}{{\sqrt 2 }} + \int {\frac{{dt}}{{2t}}} \\ I = - \ln \left| {x - 1} \right| + \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\frac{x}{{\sqrt 2 }} + \frac{1}{2}\ln \left| {{x^2} + 2} \right| + C \\\end{aligned} \end{equation} $$
