Question

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

Answer 1

Appropriate Question :-

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

$\large\underline{\sf{Solution-}}$

Given that,

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

can be rewritten as

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

$\color{green}\rm \: \displaystyle \rm f(x)=\int {\dfrac{{e}^{3x} \: dx}{{e}^{4x}+8{e}^{2x} + 4}} - - - (1)$

Also, given that

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

can be rewritten as

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

$\color{green}\rm \: \displaystyle \rm g(x)=\int {\dfrac{{e}^{x} \: dx}{{e}^{4x}+8{e}^{2x} + 4}} - - - (2)$

Now, Consider

$\displaystyle \rm {f(x)-2g(x)}$

On substituting the values, we get

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

can be rewritten as

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

Now, we use method of Substitution to evaluate this integral.

So, Substitute

$\rm \: {e}^{x} = y\rm\implies \:{e}^{x}dx = dy$

So, above integral can be rewritten as

$= \displaystyle \rm \int \frac{{y}^{2} - 2}{ {y}^{4} + {8y}^{2} + 4} \: dy$

can be further rewritten as on dividing numerator and denominator by y^2,

$= \displaystyle \rm \int \frac{1 - \dfrac{2}{ {y}^{2} } }{ {y}^{2} + 8 + \dfrac{4}{ {y}^{2} }} \: dy$

$\displaystyle \rm \int \frac{1 - \dfrac{2}{ {y}^{2} } }{ {y}^{2}+ \dfrac{4}{ {y}^{2}} + 8} \: dy$

$\displaystyle \rm \int \frac{1 - \dfrac{2}{ {y}^{2} } }{ {\bigg(y + \dfrac{2}{y} \bigg) }^{2} - 4 + 8} \: dy$

$\displaystyle \rm \int \frac{1 - \dfrac{2}{ {y}^{2} } }{ {\bigg(y + \dfrac{2}{y} \bigg) }^{2} + 4} \: dy$

Now, Substitute

$\rm \: y + \dfrac{2}{y} = z\rm\implies \:\bigg(1 - \dfrac{2}{ {y}^{2} }\bigg)dy = dz$

So, above expression can be rewritten as

$\rm \: = \displaystyle \rm \int \frac{dz}{ {z}^{2} + 4}$

$\rm \: = \displaystyle \rm \int \frac{dz}{ {z}^{2} + {2}^{2} }$

$\rm \: = \dfrac{1}{2} {tan}^{ - 1} \dfrac{z}{2} + c$

$\rm \: = \dfrac{1}{2} {tan}^{ - 1} \dfrac{ {y}^{2} + 2}{2y} + c$

$\rm \: = \dfrac{1}{2} {tan}^{ - 1} \dfrac{{e}^{2x} + 2}{2{e}^{x}} + c$

Now, Consider

$\displaystyle \rm {f(x) + 2g(x)}$

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

$\rm \: = \displaystyle \rm\int {\dfrac{({e}^{2x} + 2){e}^{x}dx}{e^{4x}+8e^{2x}+4}}$

Now, we use method of Substitution to evaluate this integral.

So, Substitute

$\rm \: {e}^{x} = y\rm\implies \:{e}^{x}dx = dy$

$= \displaystyle \rm \int \frac{{y}^{2} + 2}{ {y}^{4} + {8y}^{2} + 4} \: dy$

$= \displaystyle \rm \int \frac{1 + \dfrac{2}{ {y}^{2} } }{ {y}^{2} + 8 + \dfrac{4}{ {y}^{2} }} \: dy$

$\displaystyle \rm \int \frac{1 + \dfrac{2}{ {y}^{2} } }{ {\bigg(y - \dfrac{2}{y} \bigg) }^{2} + 4 + 8} \: dy$

Now, Substitute

$\rm \: y - \dfrac{2}{y} = z\rm\implies \:\bigg(1 + \dfrac{2}{ {y}^{2} }\bigg)dy = dz$

So, above expression can be rewritten as

$\rm \: = \displaystyle \rm \int \frac{dz}{ {z}^{2} + 12}$

$\rm \: = \displaystyle \rm \int \frac{dz}{ {z}^{2} + {(2 \sqrt{3})}^{2} }$

$\rm \: = \dfrac{1}{2 \sqrt{3} } {tan}^{ - 1} \dfrac{z}{2 \sqrt{3} } + c$

$\rm \: = \dfrac{1}{2 \sqrt{3} } {tan}^{ - 1} \dfrac{ {y}^{2} - 2}{2 \sqrt{3} y} + c$

$\rm \: = \dfrac{1}{2 \sqrt{3} } {tan}^{ - 1} \dfrac{{e}^{2x} - 2}{2 \sqrt{3} {e}^{x}} + c$