Question

`\begin{gathered}\ \ \begin{gathered}\begin{gathered}\displaystyle \text{Let :}\\ \\\displaystyle f(x)=\int {(dx)/(e^x+8e^(-x)+4e^(-3x))} \\ \\ \\\displaystyle \text{And} \\ \\\displaystyle g(x)=\int {(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 \`

Higher Class Question!

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Answer 1

Given that.

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

and,

`\displaystyle g(x)=\int {(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 {\frac{e {}^(2x) dx}{e^(3x)+8e^(x)+4e^(-x)}}`

Now,

`\begin{gathered} \longrightarrow \: \displaystyle \int \bigg( f(x) - 2g(x) \bigg)=\int {\frac{e {}^(2x) dx}{e^(3x)+8e^(x)+4e^(-x)}} - \int {(dx)/(e^(3x)+8e^(x)+4e^(-x))} \\ \\ \longrightarrow \ \: \displaystyle \int \bigg( f(x) - 2g(x) \bigg)=\int {\frac{(e {}^(2x) - 2) dx}{e^(3x)+8e^(x)+4e^(-x)}}\end{gathered} </p><p>`

Multiplying & Dividing by e^x,

`\longrightarrow \ \: \displaystyle \int \bigg( f(x) - 2g(x) \bigg)=\int {\frac{ {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 = \frac{dt}{e {}^(x) }`

Thus,

`\longrightarrow \ \: \displaystyle \int \bigg( f(x) - 2g(x) \bigg)=\int {\frac{ {e}^(x) (t{}^(2) - 2)}{t^(4)+8t^(2)+4} } * \frac{dt}{ {e}^(x) }`

Dividing Numerator and Denominator by t²,

`\longrightarrow \ \: \displaystyle \int \bigg( f(x) - 2g(x) \bigg)=\int {\frac{ (1 - \frac{2}{ {t}^(2) } )dt}{t^(2)+8+ \frac{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 \frac{ (1 - \frac{2}{ {t}^(2) } )dt}{(t + (2)/(t) ) {}^(2) + {2}^(2) }`

Here,

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

`\implies dy = ( 1 - \frac{2}{t {}^(2) } )`

Therefore,

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

Of the form,

`\boxed{ \boxed{ \sf \displaystyle \int \frac{dx}{ {x}^(2) + {a}^(2) } = (1)/(a)tan {}^( - 1)( (x)/(a) ) }}`

Further,

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

Answer 2

attachment

Explanation:

see the attachment