First slide

Methods of integration

Question

$\int \frac{1 - x^{2}}{x (1 - 2 x)} dx$ is equal to

Moderate

A

$x + \log x - \frac{3}{4} \log | x - 1 | + C$
B

$\frac{x}{2} + \log x + \frac{1}{4} \log | x - 1 | + C$
C

$\frac{x}{2} + \log x - \frac{3}{4} \log | 2 x - 1 | + C$
D

None of the above

Solution

Let $I = \int \frac{1 - x^{2}}{x (1 - 2 x)} dx$
Here, degree of numerator is equal to degree of denominator, so divide the numerator by denominator.
Thus,
$\frac{1 - x^{2}}{x (1 - 2 x)} = \frac{x^{2} - 1}{2 x^{2} - x} = \frac{1}{2} + \frac{\frac{1}{2} x - 1}{2 x^{2} - x}$
$∴ I = \int \frac{1}{2} dx + \int \frac{\frac{1}{2} x - 1}{(2 x^{2} - x)} dx \Rightarrow I = I_{1} + I_{2} . . . . . (i)$
Where, $I_{1} = \int \frac{1}{2} dx and I_{2} = \int \frac{\frac{1}{2} x - 1}{(2 x^{2} - x)} dx$
Now, $I_{1} = \int \frac{1}{2} dx = \frac{1}{2} x + C_{1} and I_{2} = \int \frac{\frac{1}{2} x - 1}{2 x^{2} - x} dx$
Let $\frac{\frac{1}{2} x - 1}{(2 x^{2} - x)} = \frac{A}{x} + \frac{B}{(2 x - 1)} \Rightarrow \frac{\frac{1}{2} x - 1}{(2 x^{2} - x)} = \frac{A (2 x - 1) + Bx}{x (2 x - 1)}$
$\begin{aligned} \Rightarrow \frac{1}{2} x - 1 = 2 Ax - A + Bx \\ \Rightarrow \frac{1}{2} x - 1 = x (2 A + B) - A \end{aligned}$
On comparing the coefficient of x and constant term on both sides, we get
$2 A + B = \frac{1}{2} . . . . . (ii) and - A = - 1 \Rightarrow A = 1 . . . . . (iii)$
$From Eq. (iii), 2 \times 1 + B = \frac{1}{2}$
$\begin{array}{ll} \Rightarrow B = \frac{1}{2} - 2 = \frac{- 3}{2} \\ ∴ I_{2} = \int [\frac{1}{x} - \frac{3}{2 (2 x - 1)}] dx \\ = \int \frac{1}{x} dx - \frac{3}{2} \int \frac{1}{2 x - 1} dx \\ \Rightarrow I_{2} = \log x - \frac{3}{2} \frac{\log | 2 x - 1 |}{2} + C_{2} \end{array}$
On putting the values of l₁ and I₂ in Eq. (i), we get
$I = \frac{1}{2} x + \log x - \frac{3}{4} \log | 2 x - 1 | + C (C_{1} + C_{2} = C)$

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