Use any method to evaluate the integrals in Exercises 55–66.
∫ 2^x / (2²x + 2^x - 2) dx

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First, rewrite the integral to clearly identify the expressions involved: \( \int \frac{2^x}{2^{2x} + 2^x - 2} \, dx \). Notice that \(2^{2x} = (2^x)^2\), so let’s set \( t = 2^x \) to simplify the integral.

Express \( dx \) in terms of \( dt \). Since \( t = 2^x \), take the natural logarithm derivative: \( \frac{dt}{dx} = 2^x \ln(2) = t \ln(2) \), which implies \( dx = \frac{dt}{t \ln(2)} \).

Substitute \( t \) and \( dx \) into the integral: \( \int \frac{t}{t^2 + t - 2} \cdot \frac{dt}{t \ln(2)} = \int \frac{1}{(t^2 + t - 2) \ln(2)} \, dt \). The \( t \) in numerator and denominator cancels out.

Focus on the integral \( \int \frac{1}{t^2 + t - 2} \, dt \). Factor the quadratic denominator: \( t^2 + t - 2 = (t + 2)(t - 1) \). Use partial fraction decomposition to express \( \frac{1}{(t + 2)(t - 1)} \) as \( \frac{A}{t + 2} + \frac{B}{t - 1} \).

Solve for constants \( A \) and \( B \), then integrate each term separately: \( \int \frac{A}{t + 2} \, dt + \int \frac{B}{t - 1} \, dt \). After integrating, substitute back \( t = 2^x \) to express the answer in terms of \( x \).

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