The partial fractions of frac{x^4 + 24x^2 + 2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $\frac{x^4 + 24x^2 + 28}{(x^2 + 1)^3} = \frac{A_1 x + B_1}{x^2 + 1} + \frac{A_2 x + B_2}{(x^2 + 1)^2} + \frac{A_3 x + B_3}{(x^2 + 1)^3}$.Then,

Vedclass · Accepted Answer

$\frac{1}{x^2 + 1} + \frac{22}{(x^2 + 1)^2} + \frac{5}{(x^2 + 1)^3}$

Vedclass · Answer

(A) Let $\frac{x^4 + 24x^2 + 28}{(x^2 + 1)^3} = \frac{A_1 x + B_1}{x^2 + 1} + \frac{A_2 x + B_2}{(x^2 + 1)^2} + \frac{A_3 x + B_3}{(x^2 + 1)^3}$.Then,$x^4 + 24x^2 + 28 = (A_1 x + B_1)(x^2 + 1)^2 + (A_2 x + B_2)(x^2 + 1) + (A_3 x + B_3)$.Let $x^2 = y$. Then $\frac{y^2 + 24y + 28}{(y + 1)^3} = \frac{A}{y + 1} + \frac{B}{(y + 1)^2} + \frac{C}{(y + 1)^3}$.Let $y + 1 = t$,so $y = t - 1$. Then $y^2 + 24y + 28 = (t - 1)^2 + 24(t - 1) + 28 = t^2 - 2t + 1 + 24t - 24 + 28 = t^2 + 22t + 5$.Thus,$\frac{t^2 + 22t + 5}{t^3} = \frac{1}{t} + \frac{22}{t^2} + \frac{5}{t^3}$.Substituting $t = x^2 + 1$,we get $\frac{1}{x^2 + 1} + \frac{22}{(x^2 + 1)^2} + \frac{5}{(x^2 + 1)^3}$.

The partial fractions of $\frac{x^4 + 24x^2 + 28}{(x^2 + 1)^3}$ are

Similar Questions

The fraction $\frac{x^2}{(x-a)(x-b)}$ is

If $\frac{x^2+x+1}{x^2+2x+1}=A+\frac{B}{x+1}+\frac{C}{(x+1)^2}$,then $A-B$ is equal to

If $\frac{(x+1)}{(2 x-1)(3 x+1)}=\frac{A}{(2 x-1)}+\frac{B}{(3 x+1)}$,then $16 A+9 B$ is equal to

Let $x$ be a real number and $-2 < x < 2$. When $\frac{x+1}{(x+3)(x-2)}$ is expanded in powers of $x$,then the coefficient of $x^3$ is

The coefficient of ${x^n}$ in the expansion of $\frac{1}{(1 - x)(3 - x)}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module