Problem 57 - Calculus Single Variable

Question

Calculate the integrals. $$ \int x^{5} \cdot\left(6+x^{2}\right)^{-1 / 2} d x $$

Step-by-Step Solution

Answer

\( \int x^{5} \cdot (6 + x^{2})^{-1/2} \, dx \) results in \( \frac{1}{5} (6 + x^2)^{5/2} - 4(6 + x^2)^{3/2} + 36(6 + x^2)^{1/2} + C \).

Step 1: Identify the Type of Integral

The integral \( \int x^{5} \cdot (6+x^{2})^{-1/2} dx \) involves a rational function and a square root. This suggests that substitution might be an effective method to simplify the integral.

Step 2: Choose a Substitution

Let's choose the substitution \( u = 6 + x^2 \). This simplifies the expression inside the square root. Consequently, the differential \( du \) of \( u \) is \( du = 2x \, dx \) or \( x \, dx = \frac{1}{2} du \).

Step 3: Express in Terms of New Variable

Using the substitution \( u = 6 + x^2 \), we can rearrange for \( x^2 = u - 6 \). So, \( x^4 = (x^2)^2 = (u-6)^2 \) and \( dx = \frac{du}{2x} \). The integral in terms of \( u \) becomes:\[ \int x^{4} \frac{1}{\sqrt{u}} \cdot \frac{1}{2x} du = \frac{1}{2} \int \frac{(u-6)^2}{\sqrt{u}} du \]

Step 4: Simplify the Integral

Expand \( (u - 6)^2 \) to get \( u^2 - 12u + 36 \). The integral becomes:\[ \frac{1}{2} \int \left( \frac{u^2}{\sqrt{u}} - \frac{12u}{\sqrt{u}} + \frac{36}{\sqrt{u}} \right) du = \frac{1}{2} \int (u^{3/2} - 12u^{1/2} + 36u^{-1/2}) \, du \].

Step 5: Integrate Each Term

Now integrate each term separately:- \( \int u^{3/2} \, du = \frac{2}{5} u^{5/2} + C_1 \)- \( \int 12u^{1/2} \, du = 24u^{3/2}/3 + C_2 = 8u^{3/2} + C_2 \)- \( \int 36u^{-1/2} \, du = 72 u^{1/2} + C_3 \)Putting it all together:\[ \frac{1}{2} \left( \frac{2}{5} u^{5/2} - 8u^{3/2} + 72u^{1/2} \right) \]

Step 6: Revert to Original Variable

Replace \( u \) back with the original expression in terms of \( x \), which is \( 6 + x^2 \):\[ \frac{1}{2} \left( \frac{2}{5} (6 + x^2)^{5/2} - 8(6 + x^2)^{3/2} + 72(6 + x^2)^{1/2} \right) + C \]

Step 7: Simplify the Result

Simplify the expression when possible:\[ \frac{1}{5} (6 + x^2)^{5/2} - 4(6 + x^2)^{3/2} + 36(6 + x^2)^{1/2} + C \]This is the final result for the integral.

Key Concepts

Integral Calculus

Integral calculus forms the backbone of understanding how to find areas under curves, among other applications. This branch of calculus revolves around the concept of integration, which is essentially the reverse process of differentiation.

When you encounter a function and want to derive its accumulation function, or simply put, how the function adds up over a range, you use integration. For instance, in the equation \( \int x^{5} \cdot (6+x^{2})^{-1/2} \, dx \), you are essentially finding the area accumulated by this function across the specified variable range.

Understanding integrals can help solve complex geometric and physical problems. Learning to break down and transform functions into a more simplified form to integrate is a key skill. Mastering this enables one to access a wide range of applications from calculating areas to solving physical problems.

Substitution Method

The substitution method is a powerful technique used in integral calculus to simplify integrands. It involves substituting a part of the integral with a new variable, to make the integration process manageable.

In our exercise, we substitute \( u = 6 + x^2 \), which directly influences the radical \((6 + x^2)^{-1/2}\). This substitution transforms a complex expression into a simpler one. In doing so, it simplifies the process and enhances the ability to perform basic integration operations.

Deriving \( du = 2x \, dx \) helps relate this new variable back to the original to ensure proper integration. When using this method, always remember to express every part of the integral in terms of the new variable, and finally, reverse the substitution to arrive back at the variable you started with.

Rational Functions

Rational functions are quotients of polynomial expressions, and they often appear in calculus problems. They can be somewhat intimidating to work with due to their complex nature, but strategies like substitution can help.

In our integral exercise, the expression \( (6 + x^2)^{-1/2} \) is an example of a rational and square root function. Such functions test the ability to reshape and rewrite equations in a form that is more convenient for integration.

Understanding the properties of rational functions, including how they interact with operations like division and square roots, aids in simplifying them into elementary integrable pieces. For instance, once substituted, the expression turns into easily manageable terms of \( u^{3/2} \), \( u^{1/2} \), and \( u^{-1/2} \) allowing straightforward integration.

Square Roots

Square roots can pose a challenging yet interesting aspect of calculus. They add complexity to integration due to their inherent properties, often complicating the polynomial landscape of an equation.

This can be seen in our integral \((6 + x^2)^{-1/2}\), where the square root is in the denominator, adding a layer of complexity. The square root modification in the integrand requires thoughtful approaches, like substitution, to handle effectively.

Simplifying expressions involving square roots can often be the key to unlock an integration problem. By transforming the integral into a polynomial format, one can effectively manage and integrate it. Converting expressions using identities or substitutions helps in making square roots approachable and solvable.