Problem 28 - Applied Calculus

Question

Find the integrals .Check your answers by differentiation. $$\int \frac{t}{1+3 t^{2}} d t$$

Step-by-Step Solution

Answer

The integral is \( \frac{1}{6} \ln |1 + 3t^2| + C \).

Step 1: Identify the Type of Integral and Substitute

The integral is of the form \( \int \frac{t}{1+3t^2} \, dt \). To solve it, we notice that it is suitable for a substitution. Let's use the substitution \( u = 1 + 3t^2 \).

Step 2: Differentiate the Substitution

Differentiate the substitution with respect to \( t \) to find \( du \).\[ \frac{du}{dt} = 6t \]Therefore, \( du = 6t \, dt \), or \( t \, dt = \frac{1}{6} \, du \).

Step 3: Rewrite the Integral

Substitute \( u \, ext{and} \, du \) back into the integral:\[ \int \frac{t}{1+3t^2} \, dt = \int \frac{1}{6u} \, du \]

Step 4: Integrate

The integral \( \int \frac{1}{6u} \, du \) can be integrated directly as a logarithmic function:\[ \int \frac{1}{6u} \, du = \frac{1}{6} \ln|u| + C \]

Step 5: Back Substitute the Original Variable

Replace \( u \) with the original expression \( 1 + 3t^2 \):\[ \frac{1}{6} \ln|1 + 3t^2| + C \]

Step 6: Differentiate to Check the Answer

Differentiate \( \frac{1}{6} \ln|1 + 3t^2| \) to ensure it results in the original integrand.\( \frac{d}{dt}\left(\frac{1}{6} \ln |1 + 3t^2|\right) = \frac{1}{6} \cdot \frac{1}{1+3t^2} \cdot \frac{d}{dt}(1 + 3t^2) = \frac{1}{6} \cdot \frac{6t}{1+3t^2} = \frac{t}{1+3t^2} \), which matches the original integrand.

Key Concepts

Logarithmic Integration

Logarithmic integration often appears when you encounter an integral of the form \( \int \frac{1}{u} \, du \). This is because such integrals lead to a logarithmic result. In this exercise, after making the substitution \( u = 1 + 3t^2 \), the given integral \( \int \frac{t}{1+3t^2} \, dt \) simplifies to \( \int \frac{1}{6u} \, du \). This integral fits the perfect logarithmic pattern.- Upon integration, you obtain \( \frac{1}{6} \ln|u| + C \), where \( C \) is the constant of integration.- To return to the original variable \( t \), remember to substitute back the expression for \( u \), resulting in \( \frac{1}{6} \ln|1 + 3t^2| + C \).Logarithmic integration is a powerful tool for evaluating integrals that involve rational functions where the denominator includes a polynomial expression.

Differentiation

Differentiation is a fundamental operation in calculus, used to determine the rate at which a function is changing. In this problem, we use differentiation to verify that our integrated solution is correct.- For the expression \( \frac{1}{6} \ln|1 + 3t^2| + C \), we need to verify by differentiating it.- Applying the chain rule, first differentiate the logarithmic function \( \ln|1 + 3t^2| \).- The derivative \( \frac{d}{dt}(1 + 3t^2) \) is \( 6t \), thus differentiating gives \( \frac{1}{1+3t^2} \times 6t \).- Hence, the full derivative is \( \frac{1}{6} \times \frac{6t}{1+3t^2} = \frac{t}{1+3t^2} \).This process confirms the original function under the integral is indeed the correct one.

Checking Integrals

Checking integrals is crucial for ensuring calculation accuracy. In this example, we check that our result is correct by differentiating the integrated expression.- The process includes identifying whether the differentiated form returns to the original function found inside the integral.- After obtaining \( \frac{1}{6} \ln|1 + 3t^2| + C \), its differentiation should result in \( \frac{t}{1+3t^2} \).- Understand that the checking step is crucial not just for verifying correctness, but also for reinforcing your comprehension of integration and differentiation.Remember, always perform this step to confirm your solutions in integral calculus. It helps solidify your understanding and ensures that no errors were made in the integration process.