Problem 39

Question

Find the integrals .Check your answers by differentiation. $$\int \frac{y}{y^{2}+4} d y$$

Step-by-Step Solution

Verified
Answer
The integral is \( \frac{1}{2} \ln |y^2 + 4| + C \).
1Step 1: Identify the Type of Integral
The given integral is \( \int \frac{y}{y^2+4} \, dy \), which takes the form of \( \int \frac{u'}{u} \, du \). This is suitable for substitution, where the numerator is the derivative of part of the denominator.
2Step 2: Substitution
Let \( u = y^2 + 4 \). Then \( \frac{du}{dy} = 2y \), which implies \( du = 2y \, dy \). Solve for \( y \, dy \) to match the integral: \( y \, dy = \frac{1}{2} du \).
3Step 3: Substitute and Simplify the Integral
Rewrite the integral in terms of \( u \):\[ \int \frac{y}{y^2+4} \, dy = \int \frac{1}{2} \frac{1}{u} \, du. \]This simplifies to \( \frac{1}{2} \int \frac{1}{u} \, du \).
4Step 4: Integrate
The integral \( \int \frac{1}{u} \, du \) is the natural logarithm, \( \ln |u| \). Thus, the integral becomes:\[ \frac{1}{2} \ln |u| + C, \]where \( C \) is the constant of integration.
5Step 5: Substitute Back in Terms of y
Replace \( u \) with \( y^2 + 4 \), achieving the antiderivative:\[ \frac{1}{2} \ln |y^2 + 4| + C. \]
6Step 6: Differentiate to Verify
Differentiate \( \frac{1}{2} \ln |y^2 + 4| + C \) with respect to \( y \):Apply the chain rule: the derivative of \( \ln |y^2 + 4| \) is \( \frac{1}{y^2 + 4} \cdot 2y \). Thus, the derivative is \[ \frac{1}{2} \cdot \frac{2y}{y^2 + 4} = \frac{y}{y^2 + 4}, \]which matches the original integrand.

Key Concepts

Substitution MethodIndefinite IntegralDifferentiation Verification
Substitution Method
The substitution method is an essential technique for solving integrals, especially when the integral's form allows for simplification.
This method is particularly handy when you notice that a part of the integral's denominator (or other parts) looks like it could simplify using a derivative that's present in the numerator.
  • For the given integral, \( \int \frac{y}{y^2+4} \, dy \), the integral can be transformed by identifying \( u = y^2 + 4 \).
  • The derivative, \( \frac{du}{dy} = 2y \), allows for rewriting as \( du = 2y \, dy \), which then simplifies the integral into a more manageable form of \( \int \frac{1}{2} \cdot \frac{1}{u} \, du \).
  • This change of variables (substitution) transforms the integral into a simpler form that is easier to integrate.
By using substitution, complex integrals become more accessible, enabling a direct path to integration once the substitution is identified.
Indefinite Integral
Integrating a function without specified limits is known as finding an indefinite integral.
It represents a family of functions and includes a constant of integration, \( C \), that reflects the unknown initial conditions.
  • In our example, once simplified using substitution, the integral becomes \( \frac{1}{2} \int \frac{1}{u} \, du \), which integrates to \( \frac{1}{2} \ln |u| + C \).
  • This result is an indefinite integral because it provides a general solution including \( C \), showing every potential antiderivative of the original function.
  • Instead of resulting in a single numerical value, an indefinite integral provides a general formula representing an infinite set of solutions.
Understanding indefinite integrals is vital as they form the foundational blocks for solving many calculus problems, offering solutions that describe entire families of curves.
Differentiation Verification
Differentiation verification is a powerful step to ensure the correctness of your integration result.
This process involves differentiating the antiderivative you obtained to check if it returns to the original integrand.
  • In our case, after integration, you have \( \frac{1}{2} \ln |y^2 + 4| + C \).
  • Differentiating this gives \( \frac{1}{2} \cdot \left( \frac{1}{y^2 + 4} \cdot 2y \right) = \frac{y}{y^2 + 4} \), returning us to the original function.
  • This confirms that the antiderivative is correct and reinforces the understanding that the substitution step was applied correctly.
Differentiation verification is essential in mathematics because it ensures the reliability and accuracy of calculus operations, allowing you to double-check your work and reinforce your learning.