Question

Use a table of integrals to evaluate the following indefinite integrals. Some of the integrals require preliminary work, such as completing the square or changing variables, before they can be found in a table. $$\int \frac{x}{\sqrt{4 x+1}} d x$$

Short answer

Question: Evaluate the indefinite integral: $$\int \frac{x}{\sqrt{4 x+1}} d x$$ Answer: $$\int \frac{x}{\sqrt{4 x+1}} d x = \frac{1}{16} \left(\frac{2}{3}(4x+1)^{\frac{3}{2}} - 2(4x+1)^{\frac{1}{2}}\right) + C$$

Step-by-step solution

Make a substitution

Let \(u = 4x + 1\). Then we have to find \(d u\) by differentiating \(u\) with respect to \(x\): $$\frac{d u}{d x} = 4$$ Now, solve for \(d x\): $$d x = \frac{d u}{4}$$ Now we can rewrite the integral in terms of \(u\): $$\int \frac{x}{\sqrt{4 x+1}} d x = \int \frac{x}{\sqrt{u}} \frac{d u}{4}$$ To rewrite the \(x\) term in terms of \(u\), solve for \(x\) from the substitution equation: $$x = \frac{u-1}{4}$$ Now the integral becomes: $$\int \frac{\frac{u-1}{4}}{\sqrt{u}} \frac{d u}{4}$$

Simplify the integral

Combine the constants and simplify the expression inside the integral: $$\int \frac{\frac{u-1}{4}}{\sqrt{u}} \frac{d u}{4} = \frac{1}{16} \int \frac{u-1}{\sqrt{u}} d u$$ Now, split the fraction inside the integral into two separate fractions: $$\frac{1}{16} \int \frac{u-1}{\sqrt{u}} d u = \frac{1}{16} \int \left(\frac{u}{\sqrt{u}} - \frac{1}{\sqrt{u}}\right) d u$$ Simplify the fractions further: $$\frac{1}{16} \int \left(\frac{u}{\sqrt{u}} - \frac{1}{\sqrt{u}}\right) d u = \frac{1}{16} \int \left(\sqrt{u} - u^{-\frac{1}{2}}\right) d u$$

Evaluate the integral

Integrate each term separately: $$\frac{1}{16} \int \left(\sqrt{u} - u^{-\frac{1}{2}}\right) d u = \frac{1}{16} \left(\int \sqrt{u} d u - \int u^{-\frac{1}{2}} d u\right)$$ Now we can use the power rule to integrate each term: $$\frac{1}{16} \left(\int \sqrt{u} d u - \int u^{-\frac{1}{2}} d u\right) = \frac{1}{16} \left(\frac{2}{3}u^{\frac{3}{2}} - 2u^{\frac{1}{2}}\right) + C$$ Where \(C\) is the constant of integration.

Substitute back the variable x

Now, we need to substitute back \(x\) into our solution. Recall that \(u = 4x + 1\). Replace \(u\) with this expression: $$\frac{1}{16} \left(\frac{2}{3}(4x+1)^{\frac{3}{2}} - 2(4x+1)^{\frac{1}{2}}\right) + C$$ And this is the final result of the indefinite integral evaluation: $$\int \frac{x}{\sqrt{4 x+1}} d x = \frac{1}{16} \left(\frac{2}{3}(4x+1)^{\frac{3}{2}} - 2(4x+1)^{\frac{1}{2}}\right) + C$$

Key concepts

Integration Techniques

Integration is a fundamental tool in calculus, where it serves as the reverse operation of differentiation. Integration techniques involve a range of methods for finding the integral or antiderivative of functions. Some common techniques include the power rule, substitution method, integration by parts, and partial fractions decomposition.

When approaching an integral, one of the keys is recognizing patterns and applying the appropriate technique. Substitution is often useful when an integral contains a function and its derivative. Sometimes, manipulations such as completing the square are necessary before applying a technique. In the given exercise, the substitution method simplifies the integral into a form where the power rule can be applied.

Substitution Method

The substitution method, also known as u-substitution, is one of the most widely used techniques for integration. It is particularly handy when you notice a composite function that is the product of a function and its derivative. To apply this method, you choose a part of the integral to substitute with a new variable, usually denoted as u. This simplification can change a challenging integral into a much easier one.

In the example, the substitution u = 4x + 1 was selected because the derivative of 4x is a constant, making it a prime candidate for substitution. As the derivative of u with respect to x is 4, we can express dx in terms of du, simplifying the integral significantly. Remember, it's crucial to replace all instances of the original variable with the new one to perform the integration correctly.

Completing the Square

Completing the square is a technique used to convert a quadratic polynomial into a perfect square trinomial plus or minus a constant. This method is particularly useful when dealing with quadratics within integral expressions where the direct application of integration techniques is not possible.

To complete the square for a quadratic of the form ax^2 + bx + c, one rewrites it as a(x - h)^2 + k where h and k are constants derived from the original coefficients. This form exposes the core component of the quadratic as a square, which can significantly simplify the integration process. While not directly applied in this specific problem, completing the square can be very useful when integrating functions involving squareroots or when applying the substitution method to integrals of rational functions.

Power Rule for Integration

The power rule for integration is a direct counterpart to the power rule for differentiation. It allows us to integrate any power of x easily and is a fundamental tool in solving indefinite integrals. The rule states that the integral of x^n, where n is any real number except -1, is (x^{n+1})/(n+1), plus a constant of integration.

In the provided exercise, after the substitution is made and the integral is simplified, the power rule is used to integrate u^{1/2} and u^{-1/2}. These correspond to u raised to the power of 1/2 and -1/2, respectively. Following the power rule, the integrals become u^{3/2} times 2/3 and u^{1/2} times -2, plus the constant of integration. Recognizing when and how to apply the power rule is vital for any student learning integration.