Question

\begin{aligned} &\text { Find the value of 'a' such that }\\\ &\int_{0}^{a} \frac{d x}{e^{x}+4 e^{-x}+5}=\ln \sqrt[3]{2} . \end{aligned}

Short answer

Answer: The value of 'a' is $$a = \ln(4 \sqrt[3]{2} - 4)$$

Step-by-step solution

Integrating the given expression

Integrate the given expression with respect to x: $$ \int \frac{1}{e^x + 4e^{-x} + 5} \, dx $$

Substitution Method

Substitute \(u = e^x\), and then find \(\frac{du}{dx}\): $$ \frac{du}{dx} = e^x \Rightarrow dx = \frac{du}{u} $$ Substitute \(e^x\) with \(u\) in the original expression and change \(dx\) to \(\frac{du}{u}\), we get: $$ \int \frac{1}{u + 4/u + 5} \frac{du}{u} $$

Simplify the integral

Simplify the integral and find the indefinite integral: $$ \int \frac{1}{u^2 + 5u + 4} du $$ Factor the denominator: $$ \int \frac{1}{(u + 4)(u + 1)} du $$

Apply Partial Fraction Decomposition

Use partial fraction decomposition for the expression: $$ \frac{1}{(u + 4)(u + 1)} = \frac{A}{u + 4} + \frac{B}{u + 1} $$ Solve for A and B: $$ A(u + 1) + B(u + 4) = 1 $$ By comparing coefficients, we find A and B: $$A = \frac{1}{3}, \quad B = -\frac{1}{3}$$ Now our integral becomes: $$ \int \left( \frac{1}{3(u + 4)} - \frac{1}{3(u + 1)} \right) du $$

Integrate with the substitution

Integrate the expression with respect to u: $$ \int \left( \frac{1}{3(u + 4)} - \frac{1}{3(u + 1)} \right) du = \frac{1}{3} \ln |u + 4| - \frac{1}{3} \ln |u + 1| + C $$ Substitute back \(u = e^x\): $$ \frac{1}{3} \ln |e^x + 4| - \frac{1}{3} \ln |e^x + 1| + C $$

Definite integral

Find the definite integral with the given limits from 0 to a: $$ \int_{0}^{a} \frac{dx}{e^x + 4e^{-x} + 5} = \left[\frac{1}{3} \ln |e^x + 4| - \frac{1}{3} \ln |e^x + 1| \right]_0^a $$

Solve for 'a'

Set the definite integral equal to the natural logarithm of the cube root of 2: $$ \left[\frac{1}{3} \ln |e^a + 4| - \frac{1}{3} \ln |e^a + 1| - \frac{1}{3} \ln |4| + \frac{1}{3} \ln |1| \right] = \ln \sqrt[3]{2} $$ Solve for 'a': $$ \frac{1}{3} \ln \frac{|e^a + 4|}{|e^a + 1|} - \frac{1}{3} \ln 4 = \ln \sqrt[3]{2} $$ Both sides are in natural logarithm form, therefore: $$ \frac{|e^a + 4|}{|e^a + 1|} = \frac{4 \sqrt[3]{2}}{4} $$ Solve for a: $$ e^a + 4 = 4 \sqrt[3]{2} $$ Subtract 4: $$ e^a = 4 \sqrt[3]{2} - 4 $$ Take natural logarithm of both sides: $$ a = \ln(4 \sqrt[3]{2} - 4) $$ So, the value of 'a' is: $$ a = \ln(4 \sqrt[3]{2} - 4) $$