Name: Method of Partial Fractions
Uploaded: 2019-07-19
Duration: 5 min 36 s
Description: Watch this video to understand how proper rational functions are integrated when the denominator consists of a linear or a quadratic expression.

Question

Evaluate:

$\int \frac{(1 - x^{2})}{x (1 - 2 x)} d x$

Accepted Answer

Let, $I = \int \frac{(x^{2} - 1)}{x (2 x - 1)} d x$

$= \int (\frac{1}{2} + \frac{(\frac{1}{2} x - 1)}{x (2 x - 1)}) d x$

$= \int \frac{1}{2} d x + \frac{1}{2} \int \frac{x}{x (2 x - 1)} d x - \int \frac{1}{x (2 x - 1)} d x$

$I = \frac{1}{2} x + \frac{1}{2} \times \frac{\log | 2 x - 1 |}{2} - I_{1}$ ...(i) $[∵ \int \frac{f^{'} (x)}{f (x)} d x = \log |f (x)|]$

Where $I_{1} = \int \frac{1}{x (2 x - 1)} d x$ ...(ii)

Let $\frac{1}{x (2 x - 1)} = \frac{A}{x} + \frac{B}{2 x - 1}$

$\Rightarrow A (2 x - 1) + B x = 1$

Putting $2 x - 1 = 0$ $\Rightarrow x = \frac{1}{2}$

$\Rightarrow A (0) + B (\frac{1}{2}) = 1$

$\Rightarrow B = 2$

Putting $x = 0$ ,

$\Rightarrow A (0 - 1) + B (0) = 1$

$\Rightarrow A = - 1$

From equation (ii), we get

$\frac{1}{x (2 x - 1)} = - \frac{1}{x} + \frac{2}{2 x - 1}$

$\int \frac{1}{x (2 x - 1)} d x = - \int \frac{1}{x} d x + 2 \int \frac{1}{2 x - 1} d x$

$= - \log | x | + \frac{2 \log | 2 x - 1 |}{2} + c$ $[∵ \int \frac{f^{'} (x)}{f (x)} d x = \log |f (x)|]$

$= \log | 2 x - 1 | - \log x + c$

From equation (i),

$I = \frac{1}{2} x + \frac{1}{4} \log | 2 x - 1 | - \log | 2 x - 1 | + \log x + c$

$= \frac{1}{2} x - \frac{3}{4} \log | 1 - 2 x | + \log | x | + c$

Evaluate:

$\int \frac{(1 - x^{2})}{x (1 - 2 x)} d x$

Important Questions on Integration Using Partial Fractions

Learn and Prepare for any exam you want !

Evaluate: ∫1-x2x(1-2x)dx

Important Questions on Integration Using Partial Fractions

Learn and Prepare for any exam you want !

Evaluate:

$\int \frac{(1 - x^{2})}{x (1 - 2 x)} d x$