Solving the Sum of Two Exponential Integrals

The problem asks for the sum of two given integrals, \(I_1\) and \(I_2\), which involve exponential functions. We need to evaluate each integral separately and then add the results.

The integrals are:

• \(I_1 = \int\frac{e^x dx}{e^x + e^{-x}}\)
• \(I_2 = \int\frac{dx}{e^{2x} + 1}\)

Evaluating Integral \(I_1\)

Let's evaluate \(I_1 = \int\frac{e^x dx}{e^x + e^{-x}}\). We can simplify the denominator by writing \(e^{-x}\) as \(\frac{1}{e^x}\):

\(I_1 = \int\frac{e^x dx}{e^x + \frac{1}{e^x}}\)

Combine the terms in the denominator:

\(I_1 = \int\frac{e^x dx}{\frac{(e^x)^2 + 1}{e^x}} = \int\frac{e^x dx}{\frac{e^{2x} + 1}{e^x}}\)

Multiply the numerator by the reciprocal of the denominator:

\(I_1 = \int e^x \cdot \frac{e^x}{e^{2x} + 1} dx = \int \frac{e^{2x}}{e^{2x} + 1} dx\)

Now, we can use a substitution. Let \(u = e^{2x} + 1\). Then the differential \(du\) is found by differentiating \(u\) with respect to \(x\):

\(\frac{du}{dx} = \frac{d}{dx}(e^{2x} + 1) = e^{2x} \cdot 2 + 0 = 2e^{2x}\)

So, \(du = 2e^{2x} dx\). This means \(e^{2x} dx = \frac{1}{2} du\).

Substitute \(u\) and \(du\) into the integral for \(I_1\):

\(I_1 = \int \frac{1}{u} \cdot \frac{1}{2} du = \frac{1}{2} \int \frac{1}{u} du\)

The integral of \(\frac{1}{u}\) with respect to \(u\) is \(\ln|u|\). Since \(u = e^{2x} + 1\) and \(e^{2x}\) is always positive, \(e^{2x} + 1\) is also always positive, so \(|u| = u\).

\(I_1 = \frac{1}{2} \ln(u) + c_1\)

Substitute back \(u = e^{2x} + 1\):

\(I_1 = \frac{1}{2} \ln(e^{2x} + 1) + c_1\)

Evaluating Integral \(I_2\)

Next, let's evaluate \(I_2 = \int\frac{dx}{e^{2x} + 1}\). We can use a substitution here as well. Let \(u = e^x\). Then \(du = e^x dx\). To find \(dx\) in terms of \(u\) and \(du\), we have \(dx = \frac{du}{e^x} = \frac{du}{u}\). Also, \(e^{2x} = (e^x)^2 = u^2\).

Substitute \(u\) and \(dx\) into the integral for \(I_2\):

\(I_2 = \int \frac{du/u}{u^2 + 1} = \int \frac{1}{u(u^2 + 1)} du\)

This integral requires partial fraction decomposition. We want to express \(\frac{1}{u(u^2 + 1)}\) as a sum of simpler fractions:

\(\frac{1}{u(u^2 + 1)} = \frac{A}{u} + \frac{Bu + C}{u^2 + 1}\)

Multiply both sides by \(u(u^2 + 1)\) to clear the denominators:

\(1 = A(u^2 + 1) + (Bu + C)u\)

\(1 = Au^2 + A + Bu^2 + Cu\)

\(1 = (A + B)u^2 + Cu + A\)

By comparing the coefficients of the powers of \(u\) on both sides, we get a system of equations:

• Coefficient of \(u^2\): \(A + B = 0\)
• Coefficient of \(u\): \(C = 0\)
• Constant term: \(A = 1\)

From the equations, we have \(A = 1\) and \(C = 0\). Substitute \(A = 1\) into \(A + B = 0\):

\(1 + B = 0 \implies B = -1\)

So the partial fraction decomposition is:

\(\frac{1}{u(u^2 + 1)} = \frac{1}{u} + \frac{-1u + 0}{u^2 + 1} = \frac{1}{u} - \frac{u}{u^2 + 1}\)

Now integrate this expression:

\(I_2 = \int \left(\frac{1}{u} - \frac{u}{u^2 + 1}\right) du = \int \frac{1}{u} du - \int \frac{u}{u^2 + 1} du\)

The first part is straightforward:

\(\int \frac{1}{u} du = \ln|u|\)

For the second part, \(\int \frac{u}{u^2 + 1} du\), let \(v = u^2 + 1\). Then \(dv = 2u du\), so \(u du = \frac{1}{2} dv\). The integral becomes:

\(\int \frac{1}{v} \frac{1}{2} dv = \frac{1}{2} \int \frac{1}{v} dv = \frac{1}{2} \ln|v|\)

Substitute back \(v = u^2 + 1\):

\(\frac{1}{2} \ln(u^2 + 1)\) (since \(u^2+1\) is always positive)

So, \(I_2 = \ln|u| - \frac{1}{2} \ln(u^2 + 1) + c_2\)

Substitute back \(u = e^x\):

\(I_2 = \ln|e^x| - \frac{1}{2} \ln((e^x)^2 + 1) + c_2\)

Since \(e^x > 0\), \(\ln|e^x| = \ln(e^x) = x\). And \((e^x)^2 = e^{2x}\).

\(I_2 = x - \frac{1}{2} \ln(e^{2x} + 1) + c_2\)

Calculating \(I_1 + I_2\)

Now, we add the results for \(I_1\) and \(I_2\):

\(I_1 + I_2 = \left(\frac{1}{2} \ln(e^{2x} + 1) + c_1\right) + \left(x - \frac{1}{2} \ln(e^{2x} + 1) + c_2\right)\)

Combine the terms:

\(I_1 + I_2 = \frac{1}{2} \ln(e^{2x} + 1) - \frac{1}{2} \ln(e^{2x} + 1) + x + c_1 + c_2\)

The \(\ln(e^{2x} + 1)\) terms cancel out:

\(I_1 + I_2 = x + c_1 + c_2\)

Let \(c = c_1 + c_2\), where \(c\) is the new constant of integration.

\(I_1 + I_2 = x + c\)

Thus, the sum of the two integrals is \(x + c\).

Summary of Integrals

Final Answer

The sum \(I_1 + I_2\) is \(x + c\), where \(c\) is the constant of integration.

Revision Table: Essential Integration Techniques

Additional Information: Integrating Exponential Functions

Exponential functions like \(e^x\) are fundamental in calculus. Their derivative and integral are unique:

• The derivative of \(e^x\) is \(e^x\).
• The integral of \(e^x\) is \(e^x + C\).

However, integrating expressions involving combinations of \(e^x\) and \(e^{-x}\) often requires algebraic manipulation before applying standard techniques like substitution. For example, expressions involving \(e^{nx}\) often suggest a substitution involving \(u = e^{nx}\) or \(u = e^x\).

Expressions like \(e^{2x} + 1\) appearing in the denominator often arise from trigonometric or hyperbolic substitutions in other contexts, but here, algebraic manipulation and standard substitution/partial fractions proved effective.

The sum of integrals property states that \(\int (f(x) + g(x)) dx = \int f(x) dx + \int g(x) dx\). In this problem, we were given the integrals separately and asked for their sum, which is mathematically equivalent. When calculating indefinite integrals and summing them, remember to include a single arbitrary constant of integration \(c\) in the final result, representing the sum of individual constants.