Problem 43 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the indefinite integral. $$\int \frac{x+1}{\sqrt{x}-1} d x$$

Step-by-Step Solution

Answer

The indefinite integral of the given expression is: $$\int \frac{x+1}{\sqrt{x}-1} d x = 2 \left( \frac{1}{4} (\sqrt{x} - 1)^4 + (\sqrt{x} - 1)^3 + \frac{3}{2} (\sqrt{x} - 1)^2 + (\sqrt{x} - 1) \right) + C$$

Step 1: Identify a substitution to simplify the problem

We can see that if we let \(u\) be a function of the denominator, the expression will simplify. So, let's try the following substitution: \(u = \sqrt{x} - 1\) Now, we need to find the differential \(d u\) and the differential \(d x\), and substitute them into the integral.

Step 2: Find the differential d u and d x

To find \(d u\), we first need to differentiate \(u\) with respect to \(x\): \( \frac{d u}{d x} = \frac{1}{2 \sqrt{x}} \) Now we can find \(d u\): \(d u = \frac{1}{2 \sqrt{x}} d x\) From our substitution, we also know that: \(\sqrt{x} = u + 1\) Now we can substitute the expressions for x, \(d x\), and \(d u\) back into the integral.

Step 3: Substitute and simplify

Our integral now becomes: $$\int \frac{(u + 1)^2 + 1}{u} \cdot 2(u + 1) d u$$ Simplify the integral: $$\int 2(u^2 + 2u + 1)(u + 1) d u$$

Step 4: Expand and simplify the integral

Next, we expand and combine terms: $$2 \int (u^3 + 3u^2 + 3u + 1) d u$$ Now we can integrate the resulting polynomial:

Step 5: Integrate the polynomial

The integral of the polynomial is as follows: \(2 ( \frac{1}{4} u^4 + u^3 + \frac{3}{2} u^2 + u ) + C\) Now, we need to substitute back in terms of \(x\). Recall that we had: \( u = \sqrt{x} - 1 \)

Step 6: Substitute back for x

Substitute \(u\) back in terms of \(x\): \(2 ( \frac{1}{4} (\sqrt{x} - 1)^4 + (\sqrt{x} - 1)^3 + \frac{3}{2} (\sqrt{x} - 1)^2 + (\sqrt{x} - 1) ) + C\) This is the indefinite integral of the given expression. The final answer is: $$\int \frac{x+1}{\sqrt{x}-1} d x = 2 ( \frac{1}{4} (\sqrt{x} - 1)^4 + (\sqrt{x} - 1)^3 + \frac{3}{2} (\sqrt{x} - 1)^2 + (\sqrt{x} - 1) ) + C$$

Key Concepts

U-Substitution

Understanding u-substitution is essential for solving complex integrals. It's a strategy that can make an integral more manageable by simplifying the integrand. The process involves selecting a piece of the original integral's formula to replace with a single variable, typically u. This is done to transform the integral into a simpler form that is more straightforward to evaluate.

For example, in our exercise, we chose u to be \( \sqrt{x} - 1 \) because its derivative is present in the numerator of the original integrand. This kind of observation is key to selecting the correct substitution. Once substituted, we obtained a polynomial in terms of u, which is significantly easier to integrate. Without u-substitution, our original integral with a radical and a fraction would be much more challenging to approach.

Integral Calculation

Integral calculation is a fundamental process in calculus, used to find the area under a curve or to solve differential equations, among other applications. When calculating integrals, we must consider the form of the function we're integrating. Some require certain techniques, like u-substitution, to simplify them into a more solvable state.

In the given problem, our goal was to perform an indefinite integral, which results in a function plus a constant of integration, denoted as C. After simplifying the integrand using u-substitution, the calculation becomes much more manageable. The resulting integral was a polynomial in terms of u, which we could then integrate term by term.

Algebraic Manipulation

Algebraic manipulation is a skill that includes rearranging, simplifying, and factoring algebraic expressions to facilitate solving problems, such as integration. When confronted with complicated expressions, as seen in the integral calculation we're working on, these techniques can be pivotal.

Once we applied u-substitution to the integral, we had to manipulate the resulting expression algebraically to prepare for integration. This involved expanding and simplifying terms so we could easily integrate the polynomial expression term by term. Mastering algebraic manipulation can often mean the difference between being able to solve an integral or not.

Polynomial Integration

The term polynomial integration refers to integrating polynomial functions, which are sums of terms with variables raised to whole number exponents. Polynomial functions are typically easier to integrate because the general method involves raising the exponent by one and dividing by the new exponent for each term.

For the integral in our exercise, after substituting and simplifying, we were left with a polynomial in u. We integrated each term by increasing the power by one and then dividing by that new power. This process is systematic and can be applied to each term of a polynomial, making integration much more straightforward once the integral is expressed in a polynomial form.