∫x4.ex.dx(x4+4x3+12x2+24x+72ex+24)2=f(x)g(x)+c and g(0)=96 then f(0) is

I=∫x4.e−x.dx(e−x(x4+4x3+12x2+24x+24)+72)2Let e−x(x4+4x3+12x2+24x+24)=t⇒−x4.e−x.dx=dt ∴I=∫−dt(t+72)2=1t+72+c =exx4+4x3+12x2+24x+24+72.ex+c g(0)=96⇒f(0)=1

∫x4.ex.dx(x4+4x3+12x2+24x+72ex+24)2=f(x)g(x)+c and g(0)=96 then f(0) is

I=∫x4.e−x.dx(e−x(x4+4x3+12x2+24x+24)+72)2Let e−x(x4+4x3+12x2+24x+24)=t⇒−x4.e−x.dx=dt ∴I=∫−dt(t+72)2=1t+72+c =exx4+4x3+12x2+24x+24+72.ex+c g(0)=96⇒f(0)=1

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$\int \frac{x^{4} . e^{x} . dx}{{(x^{4} + 4 x^{3} + 12 x^{2} + 24 x + 72 e^{x} + 24)}^{2}} = \frac{f (x)}{g (x)} + c$ and g(0)=96 then f(0) is

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$- \frac{1}{2}$

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Detailed Solution

$I = \int \frac{x^{4} . e^{- x} . dx}{{(e^{- x} (x^{4} + 4 x^{3} + 12 x^{2} + 24 x + 24) + 72)}^{2}}$
Let $e^{- x} (x^{4} + 4 x^{3} + 12 x^{2} + 24 x + 24) = t$
$\Rightarrow - x^{4} . e^{- x} . dx = dt ∴ I = \int \frac{- dt}{{(t + 72)}^{2}} = \frac{1}{t + 72} + c = \frac{e^{x}}{x^{4} + 4 x^{3} + 12 x^{2} + 24 x + 24 + 72 . e^{x}} + c g (0) = 96 \Rightarrow f (0) = 1$