Question

Evaluate the following integrals. $$\int \frac{e^{x}}{e^{2 x}+2 e^{x}+1} d x$$

Short answer

Question: Evaluate the integral $\int \frac{e^x}{e^{2x}+2e^x+1} dx$. Answer: The integral evaluates to $-\frac{1}{e^x+1}+C$.

Step-by-step solution

Rewrite the integrand in terms of partial fractions

We will now rewrite the integrand in terms of partial fractions: $$\frac{1}{(u+1)^2} = \frac{A}{u+1} + \frac{B}{(u+1)^2}$$ To find \(A\) and \(B\), we clear the fractions by multiplying both sides of the expression by \((u+1)^2\): $$1 = A(u+1)+B$$ Now, we will find the constants \(A\) and \(B\) by substituting appropriate values of \(u\) to eliminate each term.

Find the constants A and B

First, let us find \(A\). To eliminate the term with \(B\), we choose \(u=-1\): $$1=A(-1+1)+B(-1+1)^2 \implies A(0)+B(0)=1$$ This yields no information about \(A\) or \(B\), so we need to take the derivative of both sides with respect to \(u\): $$ 0 = A + 2B(u+1)$$ Now, we can try plugging in \(u=-1\): $$0 = A+2B(-1+1) \implies A = 0$$ Now let's find \(B\). Substitute \(u = 0\) in the equation for \(A\) and \(B\): $$1 = A(0+1) + B \implies B=1$$ So we have \(A=0\) and \(B=1\). Now we can rewrite the integrand in terms of partial fractions: $$\frac{1}{(u+1)^2} = \frac{0}{u+1} + \frac{1}{(u+1)^2}$$ Now the integral becomes: $$\int \frac{1}{(u+1)^2} du = \int \left(\frac{0}{u+1} + \frac{1}{(u+1)^2}\right) du$$

Integrate the partial fractions

We can integrate each term separately: $$\int \left(\frac{0}{u+1} + \frac{1}{(u+1)^2}\right) du = 0\int \frac{1}{u+1} du + \int \frac{1}{(u+1)^2} du$$ We now integrate each term: $$0\ln|u+1|-\frac{1}{u+1}+C$$

Substitute back using the original variable \(x\)

Recall that \(u = e^x\). Replace every instance of \(u\) with \(e^x\): $$-\frac{1}{e^x+1}+C$$ So the final solution to the given integral is: $$\int \frac{e^x}{e^{2x}+2e^x+1} dx = -\frac{1}{e^x+1}+C$$

Key concepts

Partial Fractions Method

The Partial Fractions Method is a powerful algebraic tool used in calculus to simplify the process of integrating complex rational expressions. It involves breaking down a complicated fraction into simpler fractions that are easier to integrate. Imagine you're trying to eat a pizza that's too big to fit in your mouth. You'd naturally slice it into more manageable pieces, and partial fractions do a similar thing for integrals.

To perform partial fraction decomposition, the denominator of a rational expression is factored into simpler terms, and then the original expression is rewritten as a sum of fractions, each with one of these simpler factors in the denominator. In the context of the given exercise, where we originally have an integral in the form of \(\int \frac{e^{x}}{e^{2x}+2e^{x}+1} dx\), we first let \(u = e^{x}\) to express the integrand as a rational function of \(u\).

Afterward, we establish a formula in terms of \(A\) and \(B\), which are constants to be determined. We look for a combination of \(\frac{A}{u+1}\) and \(\frac{B}{(u+1)^2}\) that equals the transformed function. By equating coefficients or strategically choosing values for \(u\), we solve for these constants and thus create a set of simpler integrals. In the exercise provided, determining values for \(A\) and \(B\) enabled the fraction to be decomposed effectively, leading to an integral that was manageable and straightforward to solve.

Integration Techniques

When it comes to Integration Techniques, there are several strategies mathematicians have developed to tackle challenging integrals. Just like a mechanic has a toolbox for fixing cars, mathematicians have different integration techniques for solving diverse integral problems. The exercise provided is an excellent example of blending multiple techniques to simplify and solve an integral.

After using partial fractions, we end up with a sum of simpler integrals that we can tackle individually. The sum \(\int \frac{1}{(u+1)^{2}} du\) may look intimidating at first glance, but knowing the right technique simplifies the task. For instance, recognizing how to integrate functions in the form of \(\frac{1}{(u+a)^{n}}\) where \(n > 1\) and \(a\) is a constant is vital.

One technique for such integrals is to apply the reverse power rule, simplifying \(\int \frac{1}{(u+1)^{2}} du\) to \(\frac{-1}{u+1} + C\) where \(C\) is the integration constant. This technique, alongside substitution and partial fraction decomposition, forms a multi-step approach that was beautifully demonstrated in the step-by-step solution. Understanding and mastering a variety of integration techniques can significantly boost one's ability to solve complex calculus problems.

Exponential Functions Integration

Integration of Exponential Functions often seems tough because exponential expressions can climb rapidly, seeming to defy straightforward integration. But, like a smooth dance, once you know the steps, it becomes gracefully simple. Exponential functions, such as \(e^{x}\), have the distinct feature that their rate of growth is directly proportional to their value, which makes their integration quite elegant.

When integrating exponential functions, we often look for ways to simplify the expression, and one such method is substitution, which was used in our exercise. By letting \(u = e^{x}\), we reduced the original problem to a form that was easier to manage using partial fractions and then integrated as we would with a polynomial.

The integral \(\int \frac{e^{x}}{e^{2x}+2e^{x}+1} dx\) might look daunting, but after the substitution, it significantly simplifies. The properties of the exponential function—especially that \(\frac{d}{dx}e^{x} = e^{x}\)—are precisely what make it manageable. The final substitution step brings us full circle, elegantly linking the abstract substitution back to the original variable. By recognizing these exponential properties, integrating functions like these becomes not only feasible but also quite systematic.