Question

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

Answer:

attachment

Step-by-step explanation:

see the attachment

Answer 2

Given that.

$\displaystyle f(x)=\int {\dfrac{dx}{e^x+8e^{-x}+4e^{-3x}}}$

and,

$\displaystyle g(x)=\int {\dfrac{dx}{e^{3x}+8e^{x}+4e^{-x}}}$

To perform mathematical operations on f(x) and g(x), they should have the same denominator.

Multiplying & Dividing f(x) by e^2x,

$\longrightarrow \: \displaystyle f(x)=\int {\dfrac{e {}^{2x} dx}{e^{3x}+8e^{x}+4e^{-x}}}$

Now,

$\begin{gathered} \longrightarrow \: \displaystyle \int \bigg( f(x) - 2g(x) \bigg)=\int {\dfrac{e {}^{2x} dx}{e^{3x}+8e^{x}+4e^{-x}}} - \int {\dfrac{dx}{e^{3x}+8e^{x}+4e^{-x}}} \\ \\ \longrightarrow \ \: \displaystyle \int \bigg( f(x) - 2g(x) \bigg)=\int {\dfrac{(e {}^{2x} - 2) dx}{e^{3x}+8e^{x}+4e^{-x}}}\end{gathered}

$

Multiplying & Dividing by e^x,

$\longrightarrow \ \: \displaystyle \int \bigg( f(x) - 2g(x) \bigg)=\int {\dfrac{ {e}^{x} (e {}^{2x} - 2) dx}{e^{4x}+8e^{2x}+4} }$

Let's assume t = e^x

Differentiating on both sides w.r.t x,

$dx = \dfrac{dt}{e {}^{x} }$

Thus,

$\longrightarrow \ \: \displaystyle \int \bigg( f(x) - 2g(x) \bigg)=\int {\dfrac{ {e}^{x} (t{}^{2} - 2)}{t^{4}+8t^{2}+4} } \times \dfrac{dt}{ {e}^{x} }$

Dividing Numerator and Denominator by t²,

$\longrightarrow \ \: \displaystyle \int \bigg( f(x) - 2g(x) \bigg)=\int {\dfrac{ (1 - \dfrac{2}{ {t}^{2} } )dt}{t^{2}+8+ \dfrac{4}{ {t}^{2} } } }$

Consider t² + 8 + 4/t².

It can be rewritten as (t + 2/t)² + 4 = (t + 2/t)² + 2

$\longrightarrow \ \: \displaystyle \int \bigg( f(x) - 2g(x) \bigg)=\int \dfrac{ (1 - \dfrac{2}{ {t}^{2} } )dt}{(t + \dfrac{2}{t} ) {}^{2} + {2}^{2} }$

Here,

If y = t + 2/t and differentiating w.r.t t,

$\implies dy = ( 1 - \dfrac{2}{t {}^{2} } )$

Therefore,

$\longrightarrow \ \: \displaystyle \int \bigg( f(x) - 2g(x) \bigg)=\int \dfrac{dy}{ {y}^{2} + {2}^{2} }$

Of the form,

$\boxed{ \boxed{ \sf \displaystyle \int \dfrac{dx}{ {x}^{2} + {a}^{2} } = \dfrac{1}{a}tan {}^{ - 1}( \dfrac{x}{a} ) }}$

Further,

$\begin{gathered} \longrightarrow \ \: \displaystyle \int \bigg( f(x) - 2g(x) \bigg) = \dfrac{1}{2} {tan}^{ - 1} \bigg( \dfrac{y}{2} \bigg) \\ \\ \longrightarrow \ \: \displaystyle \int \bigg( f(x) - 2g(x) \bigg) = \dfrac{1}{2} {tan}^{ - 1} \bigg( \dfrac{t {}^{2} + 2}{2t} \bigg) \\ \\ \longrightarrow \ \boxed{ \boxed{\displaystyle \int \bigg( f(x) - 2g(x) \bigg) = \dfrac{1}{2} {tan}^{ - 1} \bigg( \dfrac{ {e}^{2x} + 2 }{2 {e}^{x} } \bigg)}}\end{gathered}$