Problem 62 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int \frac{x^{2}+2 x+4}{\sqrt{x^{2}-4 x}} d x, x>4$$

Step-by-Step Solution

Answer

Question: Find the integral of the function $$\int \frac{x^{2}+2 x+4}{\sqrt{x^{2}-4 x}} d x$$ Answer: $$\int \frac{x^{2}+2 x+4}{\sqrt{x^{2}-4 x}} d x = \frac{2}{3} (x^{2}-4x)^{\frac{3}{2}} + 16(x^{2}-4x)^{\frac{1}{2}} + C$$

Step 1: Complete the square in the numerator

We have to complete the square for the polynomial \(x^2 + 2x + 4\). To do that, we use the formula \(x^2 + 2bx + (b^2 + c) = (x + b)^2 + c - b^2\), in which b is half of the coefficient of the linear term (x) in the original expression. $$x^2 + 2x + 4 = x^2 + 2(1)x + (1^2 + 3) = (x + 1)^2 + 3$$ Now our integral becomes: $$\int \frac{(x+1)^{2}+3}{\sqrt{x^{2}-4 x}} d x$$

Step 2: Use substitution

We use the substitution \(u = x^{2}-4x\), which rewrites the integral in terms of \(u\). Let's find \(du\): $$du = \frac{d}{dx}(x^{2}-4x) = (2x - 4)dx$$ Solve for \(dx\): $$dx = \frac{du}{2x-4}$$ Replace \(x^2 - 4x\) by \(u\) and \(dx\) with its expression in terms of \(du\) in the integral: $$\int \frac{(x+1)^{2}+3}{\sqrt{u}}\frac{du}{2x-4}$$ To finish substitution, notice that when \(u = x^2 - 4x\), the expression \((x+1)^2 + 3\) can be rewritten as \(u+8\). So, the integral becomes: $$\int \frac{u+8}{\sqrt{u}}\frac{du}{2x-4}$$

Step 3: Simplify the integral

First, cancel a factor of \((2x-4)\) in the numerator and denominator: $$\int \frac{\sqrt{u}(u+8)}{u} du$$ Now, distribute the \(\sqrt{u}\) in the numerator: $$\int \frac{u\sqrt{u} + 8\sqrt{u}}{u} du = \int (u^{\frac{1}{2}} +\frac{8}{u^{\frac{1}{2}}}) du$$

Step 4: Integrate

Now integrate each term separately: $$\int u^{\frac{1}{2}} du + \int \frac{8}{u^{\frac{1}{2}}} du = \frac{2}{3} u^{\frac{3}{2}} + 16u^{\frac{1}{2}} + C$$

Step 5: Substitute back x

Replace \(u\) back with \(x^{2}-4x\): $$\frac{2}{3} (x^{2}-4x)^{\frac{3}{2}} + 16(x^{2}-4x)^{\frac{1}{2}} + C$$ This is our final result: $$\int \frac{x^{2}+2 x+4}{\sqrt{x^{2}-4 x}} d x = \frac{2}{3} (x^{2}-4x)^{\frac{3}{2}} + 16(x^{2}-4x)^{\frac{1}{2}} + C$$

Key Concepts

Completing the square

Completing the square is a valuable algebraic technique used to transform a quadratic expression into a perfect square plus or minus a constant. This transformation is often helpful for integration, particularly when dealing with quadratic expressions in the integrand. In our context, the expression \(x^2 + 2x + 4\) was transformed.
To complete the square, we adjust the quadratic expression, typically of the form \(ax^2 + bx + c\). The goal is to rewrite it as \((x + b)^2 + c - b^2\), where \(b\) is half the linear term coefficient.
In this exercise, the quadratic expression \(x^2 + 2x + 4\) becomes \((x + 1)^2 + 3\) by completing the square, allowing us to simplify the integral in subsequent steps.

Substitution method for integrals

The substitution method is a powerful technique for evaluating complex integrals by changing variables, simplifying the integration process. We use this method when substituting a part of the integrand with a new variable.
In this exercise, we began by setting \(u = x^2 - 4x\), a common practice for breaking down difficult terms (often rooted in squares) into simpler forms. By calculating the derivative, \(du = (2x - 4)dx\), we then solve for \(dx\).
This substitution reforms the integral into an easier expression concerning \(u\). It's crucial because it changes the domain from \(x\) to \(u\), hugely simplifying the process of integration.

Simplifying integrals

Simplifying integrals involves algebraic manipulation to reduce the complexity of the integral before attempting to integrate.
In the solution, this entailed canceling out common terms and redistributing factors within the integral. By rewriting the expression \((x+1)^2 + 3\) in terms of \(u\), simplifying it to \( \int \frac{u+8}{\sqrt{u}} \frac{du}{2x-4} \), and canceling the redundant terms, we focused our expression.
Eventually, it becomes \( \int (u^{\frac{1}{2}} + \frac{8}{u^{\frac{1}{2}}}) du \). This simplification made the integral easier to handle when we performed the actual integration.

Definite integrals

Definite integrals are used to compute the area under a curve from a starting point to an endpoint — typically from \(a\) to \(b\). While our problem was handling an indefinite integral, many of the techniques are shared between both.
The process and simplification methods remain consistent: complete the square, substitute when necessary, simplify where you can, and finally, solve the integral. When solving definite integrals, after evaluation, boundaries are simply applied to the final simplified expression.
Understanding the integration of definite integrals is essential because these concepts are fundamental in the computation of other integral-related calculus applications like finding volumes and solving other physical or engineering problems.