Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the indicated indefinite integrals. $$ \int \frac{\left(z^{2}+1\right)^{2}}{\sqrt{z}} d z $$

Step-by-Step Solution

Answer

\(\int \frac{(z^{2}+1)^{2}}{\sqrt{z}} \, dz = \frac{2}{9}z^{\frac{9}{2}} + \frac{4}{5}z^{\frac{5}{2}} + 2z^{\frac{1}{2}} + C\)

Step 1: Simplify the Integrand

First, let's simplify the integrand \( \frac{(z^2 + 1)^2}{\sqrt{z}} \). This can be rewritten as \( (z^2 + 1)^2 \times z^{-\frac{1}{2}} \). This will allow us to expand the expression easier.

Step 2: Expand the Polynomial

Next, expand \((z^2 + 1)^2\) by using the binomial theorem: \( (z^2 + 1)^2 = z^4 + 2z^2 + 1 \). Now our integrand becomes \((z^4 + 2z^2 + 1)z^{-\frac{1}{2}}\).

Step 3: Simplify Further

Distribute \( z^{-\frac{1}{2}} \) over each term of the polynomial: \( z^4 \cdot z^{-\frac{1}{2}} = z^{\frac{7}{2}} \), \( 2z^2 \cdot z^{-\frac{1}{2}} = 2z^{\frac{3}{2}} \), \( 1 \cdot z^{-\frac{1}{2}} = z^{-\frac{1}{2}} \). So the integrand is now \( z^{\frac{7}{2}} + 2z^{\frac{3}{2}} + z^{-\frac{1}{2}} \).

Step 4: Integrate Term by Term

Integrate each term separately:- \(\int z^{\frac{7}{2}} \, dz = \frac{z^{\frac{9}{2}}}{\frac{9}{2}} = \frac{2}{9}z^{\frac{9}{2}}\)- \(\int 2z^{\frac{3}{2}} \, dz = 2 \times \frac{z^{\frac{5}{2}}}{\frac{5}{2}} = \frac{4}{5}z^{\frac{5}{2}}\)- \(\int z^{-\frac{1}{2}} \, dz = \frac{z^{\frac{1}{2}}}{\frac{1}{2}} = 2z^{\frac{1}{2}}\)

Step 5: Combine and Add Constant of Integration

Combine the integrated terms and add the constant of integration \(C\):\[\frac{2}{9}z^{\frac{9}{2}} + \frac{4}{5}z^{\frac{5}{2}} + 2z^{\frac{1}{2}} + C\]

Key Concepts

Integration Techniques

When we talk about integration techniques, we are referring to various methods utilized to solve integrals, especially those that are not immediately straightforward. One approach for complex integrands is simplification—breaking down a troublesome expression into something more manageable.
When faced with the function \( \frac{(z^2 + 1)^2}{\sqrt{z}} \), simplifying the expression is key. Rewriting it as \( (z^2 + 1)^2 \times z^{-\frac{1}{2}} \) transforms the fraction into a multiplication, which is generally easier to handle. From here, we expand and distribute, reducing the complexity before applying integration rules. This technique enables us to integrate more challenging expressions in a step-by-step manner.

Polynomial Expansion

Polynomial expansion is a critical step in simplifying functions for integration, especially when dealing with binomials raised to a power. For instance, expanding \((z^2 + 1)^2\) involves applying the binomial theorem, resulting in \(z^4 + 2z^2 + 1\).
This step not only simplifies the expression but also prepares each term for individual integration. As each polynomial term is isolated, it can be integrated one by one, allowing the use of basic integration techniques, reducing potential errors. Therefore, mastering polynomial expansion is indispensable for tackling more complex indefinite integrals.

Power Rule in Integration

The power rule in integration is a foundational tool used to integrate functions of the form \( x^n \). This rule states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), provided that \( n eq -1 \).
Applying this rule to the integrands obtained after simplification, such as \( z^{\frac{7}{2}} \), \( 2z^{\frac{3}{2}} \), and \( z^{-\frac{1}{2}} \), allows us to find their indefinite integrals individually:

• \( \int z^{\frac{7}{2}} \, dz = \frac{z^{\frac{9}{2}}}{\frac{9}{2}} = \frac{2}{9}z^{\frac{9}{2}} \)
• \( \int 2z^{\frac{3}{2}} \, dz = 2 \cdot \frac{z^{\frac{5}{2}}}{\frac{5}{2}} = \frac{4}{5}z^{\frac{5}{2}} \)
• \( \int z^{-\frac{1}{2}} \, dz = \frac{z^{\frac{1}{2}}}{\frac{1}{2}} = 2z^{\frac{1}{2}} \)

Understanding and applying the power rule simplifies the integration process and ensures accurate results.