Saved Bookmarks
| 1. |
Evaluate : \( \int \frac{1-x^{2}}{x-2 x^{2}} d x \) |
|
Answer» Let I be the given integral \(I=\int \frac {1-x^2}{x-2x^2} dx \) \(I=\int \frac {1-x^2}{x(1-2x)} dx\) This can be further solved by the method of partial fractions Let \(\frac{1-x^2}{x(1-2x)}\) be of the form \(\frac{A}{x} + \frac{Bx+C}{1-2x}\) where A,B and C are integers By taking the LCM and by comparing both the sides, we get A=1, B=-1,C=2 Now, the given integral is simplified and is written as, \(I=\int \frac{1}{x} dx + \int \frac {-x+2}{1-2x} dx\) \(I=log(x) + \int \frac{-x+2}{1-2x} dx\) The second integral should be written in the form n(Denominator) + m, where m and n are real numbers \(\therefore I=log(x) + 1/2 \int \frac {-2x+1+3}{-2x+1} dx\) \(I=log(x) +\int \frac {dx}{2} + \int \frac {3}{2(-2x+1)} dx\) \(I= log(x) + \frac {x}{2} - \frac{3}{4} log(1-2x) + c\) Hence the value of the given integral is \(I= \frac {x}{2} + log(x) - \frac {3}{4} log(1-2x) + c\) |
|