If $\frac{1-x+6x^2}{x-x^3} = \frac{A}{x} + \frac{B}{1-x} + \frac{C}{1+x}$,then $A$ is equal to
- A$1$
- B$2$
- C$3$
- D$4$
Explore More
Similar Questions
Resolve the following expression into partial fractions: $\frac{x + 1}{(x - 1)(x - 2)(x - 3)}$
If $\frac{3x+5}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$ and $f(x) = Ax^3 + Bx^2 + 7x + C$, then $5C - f'(-2) = $
The partial fractions of $\frac{x^2 - 5}{x^2 - 3x + 2}$ are
Easy
View SolutionThe absolute value of the difference of the coefficients of $x^4$ and $x^6$ in the expansion of $\frac{2 x^2}{(x^2+1)(x^2+2)}$ is
If $\frac{2 x^2+5 x+6}{(x+2)^3}=\frac{a}{x+2}+\frac{b}{(x+2)^2}+\frac{c}{(x+2)^3}$ then $a \cdot b+b \cdot c+c \cdot a=$
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo