Number of Partial Fractions of 3x2-1-x2-14

The correct option is C 2 3x^2+1/(x^2+1)=Ax+B/x^2+1+Cx+D/(x^2+1)^2+ex+f/(x^2+1)^3+9x+4/(x^2+1)^4 LHS=RHS 3x^2+1=(Ax+B)[x^2+1]^3+(Cx+D)(x^2+1)^2 +(e ...

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Byju's Answer

Standard XII

Mathematics

Differentiability in an Interval

Number of Par...

Question

Number of Partial Fractions of $\frac{3 x^{2} + 1}{(x^{2} + 1)^{4}}$

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Solution

The correct option is C 2
$\frac{3 x^{2} + 1}{(x^{2} + 1)} = \frac{A x + B}{x^{2} + 1} + \frac{C x + D}{(x^{2} + 1)^{2}} + \frac{e x + f}{(x^{2} + 1)^{3}} + \frac{9 x + 4}{(x^{2} + 1)^{4}}$
LHS=RHS
$3 x^{2} + 1 = (A x + B) [x^{2} + 1]^{3} + (C x + D) (x^{2} + 1)^{2}$
$+ (e x + f) (x^{2} + 1)$
$+ (9 x + 4)$
comparing co-efficients of
$x^{7}, x^{6}, x^{5}, x^{4}, x^{3}$ in LHS & RHS
$A = B = C = D = e = o$
So $f (x^{2} + 1) + (9 x + 4)$ is the Use
$\frac{3 x^{2} + 1}{(x^{2} + 1)^{4}} = \frac{f}{(x^{2} + 1)^{3}} + \frac{9 x + 4}{(x^{2} + 1)^{4}}$
So, two partial fractions.

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Similar questions

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Number of Partial Fractions of 3x2+1(x2+1)4

The correct option is C 23x2+1(x2+1)=Ax+Bx2+1+Cx+D(x2+1)2+ex+f(x2+1)3+9x+4(x2+1)4LHS=RHS3x2+1=(Ax+B)[x2+1]3+(Cx+D)(x2+1)2+(ex+f)(x2+1)+(9x+4)comparing co-efficients ofx7,x6,x5,x4,x3 in LHS & RHSA=B=C=D=e=oSo f(x2+1)+(9x+4) is the Use3x2+1(x2+1)4=f(x2+1)3+9x+4(x2+1)4So, two partial fractions.

Number of Partial Fractions of $\frac{3 x^{2} + 1}{(x^{2} + 1)^{4}}$