Question

$\ \ \begin{gathered}\begin{gathered}\displaystyle \text{Let :}\\ \\\displaystyle f(x)=\int {\dfrac{dx}{e^x+8e^{-x}+4e^{-3x}}} \\ \\ \\\displaystyle \text{And} \\ \\\displaystyle g(x)=\int {\dfrac{dx}{e^{3x}+8e^{x}+4e^{-x}}} \\ \\ \\\displaystyle \text{Then find value of :} \\ \\\displaystyle \int {(f(x)-2g(x))dx} \, \quad \text{And} \\ \\ \\\displaystyle \int {(f(x)+2g(x))dx}\end{gathered}\end{gathered} \ \textless \ br /\ \textgreater$
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Answer 1

Answer:

f′(x) = lim f(x + h) − f(x) h→0h

=lim1 2 −2 h→0 h (x + h)2 x2

1 2x2 2(x+h)2 =lim2 2−2 2

h→0 h x (x+h) x (x+h) = lim 1 2x2 −2(x+h)2

h→0 h x2(x + h)2

= lim 1 2x2 −2x2 −4xh−2h2

h→0 h x2(x + h)2 = lim 1 −4xh−2h2

h→0 h x2(x + h)2 =lim −4x−2h

h→0 x2(x + h)2

= −4x − 2 ∗ 0 = −4x = −4

x2(x + 0)2 x4 x3.

Answer 2

$\bold\red{Answer}$

f′(x) = lim f(x + h) − f(x) h→0h

=lim1 2 −2 h→0 h (x + h)2 x2

1 2x2 2(x+h)2 =lim2 2−2 2

h→0 h x (x+h) x (x+h) = lim 1 2x2 −2(x+h)2

h→0 h x2(x + h)2

= lim 1 2x2 −2x2 −4xh−2h2

h→0 h x2(x + h)2 = lim 1 −4xh−2h2

h→0 h x2(x + h)2 =lim −4x−2h

h→0 x2(x + h)2

= −4x − 2 ∗ 0 = −4x = −4

x2(x + 0)2 x4 x3.

$\bold\green{Sumit \: Here \: Now}$