Question

\(\int\left[\mathrm{dx} /\left(\mathrm{e}^{\mathrm{x}}+1\right)\right]=\underline{\mathrm{c}}\) (a) \(-\log \mid\left[\left(\mathrm{e}^{\mathrm{x}}+1\right) /\left(\mathrm{e}^{\mathrm{x}}\right)\right]\) (b) \(-\log \mid\left[\left(\mathrm{e}^{\mathrm{x}}\right) /\left(\mathrm{e}^{\mathrm{x}}+1\right)\right]\) (c) \(\log \left[\left[\left(\mathrm{e}^{\mathrm{x}}+1\right) /\left(2 \mathrm{e}^{\mathrm{x}}\right)\right]\right.\) (d) \(\log \mid\left[(\mathrm{e})^{2 \mathrm{x}} /\left(\mathrm{e}^{\mathrm{x}}+1\right)\right]\)

Short answer

-\ln\left|\frac{e^x}{e^x+1}\right|

Step-by-step solution

Identify a substitution

Let's try the substitution \(u = e^x\). To do this, we must also find the derivative of \(u\) w.r.t \(x\), which is \(du = e^x dx\).

Replace variables in the integral

Using our substitution, we can rewrite the integral as follows: \[\int\frac{dx}{e^x+1} = \int\frac{du}{u(u+1)}\] since \(e^x dx = du\).

Perform partial fraction decomposition

We will now perform partial fraction decomposition on the integrand: \[\frac{1}{u(u+1)} = \frac{A}{u} + \frac{B}{u+1}\] Multiplying both sides by \(u(u+1)\) gives us: \[1=A(u+1)+Bu\] For A, let \(u=0\): \[1=A(0+1)+B(0) \Longrightarrow A=1\] For B, let \(u=-1\): \[1=A(-1+1)+B(-1)\Longrightarrow B=-1\] So our partial fraction decomposition gives us: \[\frac{1}{u(u+1)} = \frac{1}{u} - \frac{1}{u+1}\]

Integrate

Now we integrate each term separately: \[\int\frac{du}{u(u+1)} = \int\left(\frac{1}{u} - \frac{1}{u+1}\right)du = \int\frac{1}{u}du - \int\frac{1}{u+1}du\] Applying the integral to both terms, we obtain: \[\ln |u| - \ln |u+1|\]

Substitute back and simplify

Now we substitute back and obtain the antiderivative of the original integral: \[\ln |e^x| - \ln |e^x+1| = \ln\left|\frac{e^x}{e^x+1}\right|\] However, we need to add a constant of integration, so our final result is: \[\ln\left|\frac{e^x}{e^x+1}\right| + C\] The correct option is therefore (b) \( -\ln\left|\frac{e^x}{e^x+1}\right|\).

Key concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used primarily in integral calculus to simplify complex rational functions into simpler fractions that are easier to integrate. This method is particularly useful when dealing with polynomial expressions in the denominator.

Imagine you’re trying to split a complicated fraction into parts; that's essentially what partial fraction decomposition does with algebraic fractions. You're looking for numbers (constants) that, when put over their respective polynomial pieces, add up to the original fraction.

In the exercise, the integrand was a complex fraction \( \frac{1}{u(u+1)} \), where we needed to find constants A and B such that \( \frac{1}{u(u+1)} = \frac{A}{u} + \frac{B}{u+1} \). The process involved cleverly selecting values for \(u\) that simplify the equation, allowing us to solve for A and B individually. The goal was to rewrite the complex fraction into simpler fractions which could easily be integrated.

Indefinite Integral

The indefinite integral, also known as an antiderivative, is essentially the reverse operation of differentiation. If you think of differentiation as breaking down functions into smaller rates of change, integration gathers them up into a whole amount of something – like collecting rainfall over time into a measurable quantity of water.

In calculus, finding the indefinite integral of a function means identifying a function that, when differentiated, gives you the original function. Notably, because differentiation wipes out any constant term—the infamous \(C\) in the indefinite integral—the integral includes a '+ C' to represent any possible constant that was 'lost' during differentiation.

In the given exercise, after applying partial fraction decomposition, we calculated the indefinite integral of each simpler fraction, which started by integrating \( \frac{1}{u} \) and \( \frac{1}{u+1} \), leading us to natural logarithm functions.

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is the logarithm to the base \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. This mathematical function lays the groundwork for much of calculus and appears often when dealing with exponential growth or decay.

The natural logarithm has a unique property: it is the inverse operation to exponentiation with the base \(e\). In other words, if \(e^y = x\), then \(\ln(x) = y\). The natural logarithm crops up when we integrate \(\frac{1}{x}\), resulting in \(\ln |x|\).

In our problem, after integration, we obtained \(\ln |u| - \ln |u+1|\), which simplifies to \(\ln\left|\frac{e^x}{e^x+1}\right|\) using properties of logarithms. This function is fundamentally tied to rates of change and growth processes, which is why it appears in our final solution.