3x2-1-x2-13-Ax-B-x2-1-Cx-D-x2-12-Ex-F-x2-13 then A - C - E - F -

The correct option is D 2 3x^2+1(x^2+1)^3=(Ax+B)(x^2+1)^2+(Cx+D)(x^2+1)+Ex+F(x^2+1)^3 3x^2+1=(Ax+B)(x^2+1)^2+(Cx+D)(x^2+1)+(Ex+F) #160; –(1) compar ...

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Byju's Answer

Standard XII

Mathematics

Integration of Piecewise Continuous Functions

3x2+1/x2+13=A...

Question

$\frac{3 x^{2} + 1}{(x^{2} + 1)^{3}} = \frac{A x + B}{(x^{2} + 1)} + \frac{C x + D}{(x^{2} + 1)^{2}} + \frac{E x + F}{(x^{2} + 1)^{3}}$ then $A + C + E + F =$

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Solution

The correct option is D -2
$\frac{3 x^{2} + 1}{(x^{2} + 1)^{3}} = \frac{(A x + B) (x^{2} + 1)^{2} + (C x + D) (x^{2} + 1) + E x + F}{(x^{2} + 1)^{3}}$
$3 x^{2} + 1 = (A x + B) (x^{2} + 1)^{2} + (C x + D) (x^{2} + 1) + (E x + F) - - (1)$
comparing the co-efficient of $x^{5}, x^{3}, x$
we get $A = 0, C = 0; E = 0$
$3 x^{2} + 1 = B (x^{2} + 1)^{2} + D (x^{2} + 1) + F$
comparing $x^{4}$ co-efficient $B = 0$ and $x^{2}$
we get , $D = 3;$
So, $3 x^{2} + 1 = 3 (x^{2} + 1) + F$
$\Rightarrow F = - 2$

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3x2+1(x2+1)3=Ax+B(x2+1)+Cx+D(x2+1)2+Ex+F(x2+1)3 then A+C+E+F=

The correct option is D -23x2+1(x2+1)3=(Ax+B)(x2+1)2+(Cx+D)(x2+1)+Ex+F(x2+1)33x2+1=(Ax+B)(x2+1)2+(Cx+D)(x2+1)+(Ex+F)−−(1)comparing the co-efficient of x5,x3,xwe get A=0,C=0;E=03x2+1=B(x2+1)2+D(x2+1)+Fcomparing x4 co-efficient B=0 and x2we get , D=3;So, 3x2+1=3(x2+1)+F⇒F=−2

$\frac{3 x^{2} + 1}{(x^{2} + 1)^{3}} = \frac{A x + B}{(x^{2} + 1)} + \frac{C x + D}{(x^{2} + 1)^{2}} + \frac{E x + F}{(x^{2} + 1)^{3}}$ then $A + C + E + F =$