Question

integration of 1/ 4x^2 - 3x - 1 ​

Answer 1

Step-by-step explanation:

We have,

$\int \frac{1}{4 {x}^{2} - 3x - 1} dx$

$= \int \frac{1}{4 \bigg\{ {x}^{2} - \dfrac{3x}{4} - \dfrac{1}{4} \bigg \}} dx$

$= \frac{1}{4} \int \frac{1}{ {x}^{2} - \dfrac{3x}{4} - \dfrac{1}{4} } dx$

$= \frac{1}{4} \int \frac{1}{ {x}^{2} - 2. \dfrac{3}{8}.x + \dfrac{9}{64} - \dfrac{9}{64} - \dfrac{1}{4} } dx$

$= \frac{1}{4} \int \frac{1}{ {x}^{2} - 2. \dfrac{3}{8}.x + \bigg(\dfrac{3}{8} \bigg)^{2} - \dfrac{9 +16}{64} } dx$

$= \frac{1}{4} \int \frac{1}{ \bigg(x - \dfrac{3}{8} \bigg)^{2} - \dfrac{25}{64} } dx$

$= \frac{1}{4} \int \frac{1}{ \bigg(x - \dfrac{3}{8} \bigg)^{2} - \bigg(\dfrac{5}{8} \bigg)^{2} } dx$

$= \frac{1}{4} . \frac{8}{2 \times 5} \ln \left| \begin{array}{} \frac{ \large{x} - \dfrac{3}{8} - \dfrac{5}{8} }{ \large{x }- \dfrac{3}{8} + \dfrac{5}{8} } \end{array} \right| + C$

$= \frac{1}{5} \ln \bigg| \frac{ x - 1 }{ x + \dfrac{1}{4} } \bigg| + C$

$= \frac{1}{5} \ln \bigg| \frac{4 x - 4 }{ 4x + 1 } \bigg| + C$