Problem 47 - University Calculus: Early Transcendentals

Question

Evaluate the integrals. $$\int x^{3} \sqrt{x^{2}+1} d x$$

Step-by-Step Solution

Answer

\( \int x^{3} \sqrt{x^{2}+1} \, dx = \frac{1}{5}(x^2 + 1)^{5/2} - \frac{1}{3}(x^2 + 1)^{3/2} + C \)

Step 1: Choose the Substitution

In solving integrals involving polynomials and square roots, a suitable substitution often simplifies the expression. Let's use the substitution \( u = x^2 + 1 \). This leads to \( du = 2x \, dx \) or \( x \, dx = \frac{1}{2} du \).

Step 2: Adjust the Expression for Substitution

Express \( x^3 \) in terms of \( u \). Since \( u = x^2 + 1 \), we can say \( x^2 = u - 1 \), hence \( x^3 = x(x^2) = x(u-1) \). Also remember that \( x^{3} dx = x \cdot x^2 \cdot dx \).

Step 3: Substitute in Terms of u

Replace \( x^2 + 1 \) with \( u \) and \( x \, dx \) with \( \frac{1}{2} du \) in the integral. The integral becomes:\[\int x^2 \sqrt{x^2+1} \, x \, dx = \int (u-1) \sqrt{u} \cdot \frac{1}{2} du\]

Step 4: Simplify and Separate the Integral

Simplify and separate the integral:\[\frac{1}{2} \int (u-1) \sqrt{u} \, du = \frac{1}{2} \int (u^{3/2} - u^{1/2}) \, du\]

Step 5: Integrate Term by Term

Integrate each term separately:1. \( \int u^{3/2} \, du = \frac{u^{5/2}}{5/2} = \frac{2}{5} u^{5/2} \)2. \( \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2} \)Combine the results:\[\frac{1}{2} \left( \frac{2}{5} u^{5/2} - \frac{2}{3} u^{3/2} \right)\]

Step 6: Substitute Back x

Replace \( u \) back with \( x^2 + 1 \):\[= \frac{1}{2} \left( \frac{2}{5}(x^2 + 1)^{5/2} - \frac{2}{3}(x^2 + 1)^{3/2} \right)\] Simplify further to get the final result:\[= \frac{1}{5}(x^2 + 1)^{5/2} - \frac{1}{3}(x^2 + 1)^{3/2} + C\]

Step 7: Final Answer

The integral \( \int x^{3} \sqrt{x^{2}+1} \, dx \) evaluates to \[ \frac{1}{5}(x^2 + 1)^{5/2} - \frac{1}{3}(x^2 + 1)^{3/2} + C \] where \( C \) is the constant of integration.

Key Concepts

Substitution Method

One of the essential tools in integral calculus is the substitution method. This technique helps simplify complex integrals by introducing a new variable, typically noted as \( u \). In this particular exercise, we used substitution to manage the combination of a polynomial and a square root.

Here's how it works: we set \( u = x^2 + 1 \), which simplifies the expression significantly. When the original integral becomes too convoluted, a good substitution can make it easier to handle. The goal is to find a new form of the integral that is simpler to evaluate.

• Choose a substitution \( u \) that will replace part or all of the original variable expression.
• Compute \( du \), which represents the derivative of \( u \) with respect to \( x \).
• Rewrite both the integral and its differential (like \( dx \)) in terms of this new variable.

By simplifying these expressions into something more manageable, the integration process becomes straightforward.

Polynomial Integration

Polynomial integration is vital when dealing with algebraic expressions involving powers of a variable. With polynomial integration, each term of the polynomial can be integrated independently. In our example, after using substitution, we dealt with terms like \( u^{3/2} \) and \( u^{1/2} \).

For any polynomial \( ax^n \), the integral is given by:\[\int ax^n \, dx = \frac{ax^{n+1}}{n+1} + C\]This formula is incredibly powerful because it allows each component of the polynomial to be integrated term by term.

• Break the polynomial into separate terms upon which integration can be performed.
• Apply the power rule for integration to each term.
• Combine the results to build the complete integrated function.

This step-by-step approach makes polynomial integration both systematic and efficient.

Square Root Integration

Integrating functions that involve square roots can be challenging, but certain techniques often simplify the process. Typically, substitution is used to convert the square root term into something more tractable. In the example exercise, after substitution, the expression \( \sqrt{x^2 + 1} \) was simplified to \( \sqrt{u} \).

Square root terms might seem daunting because they involve powers that aren't integers. However, introducing substitution helps change the expression to a form where regular polynomial integration rules apply.

• First, use suitable substitution to handle the square root component.
• Transform the square root expression into a simple power expression.
• Apply the polynomial integration techniques to the transformed expression.

Once transformed, you can use basic rules of integration to solve these difficult integrals, bringing the entire problem within reach.