Question

Evaluate the following integrals. $$\int \ln \left(x^{2}+a^{2}\right) d x, a \neq 0$$

Short answer

Short answer: The integral of the given function is as follows: $$\int \ln \left(x^{2}+a^{2}\right) d x = x\ln(x^2 + a^2) - 2x^2 - 2a^2\ln|x^2 + a^2| + 4a^2 + C$$

Step-by-step solution

Applying Integration by Parts

We will use integration by parts to evaluate the integral. For integration by parts, we apply the formula: $$\int u dv = uv - \int v du$$ Here we take: - \(u = \ln(x^2 + a^2)\) - \(dv = dx\) Now, we find the derivatives and integrals of these components: - \(du = \frac{2x}{x^2 + a^2}dx\) - \(v = x\)

Substituting into the Integration by Parts Formula

Now we substitute these components into the integration by parts formula: $$\int \ln(x^2 + a^2)dx = x\ln(x^2 + a^2) - \int x\frac{2x}{x^2 + a^2}dx$$ Now we simplify the second integral: $$x\ln(x^2 + a^2) - 2\int \frac{x^2}{x^2 + a^2}dx$$

Using Substitution Method for the Remaining Integral

We can now use the substitution method to evaluate the remaining integral. Let's make the following substitution: $$z = x^2 + a^2$$ $$\frac{dz}{dx} = 2x$$ $$dx = \frac{dz}{2x}$$ Now we will rewrite our integral: $$2\int \frac{x^2}{z}\frac{dz}{2x} = \int \frac{z-a^2}{z}dz$$

Splitting the Integral and Integrating

Let's split the integral into two parts and find the antiderivatives: $$\int \frac{z-a^2}{z}dz = \int \left( 1 - \frac{a^2}{z} \right)dz = \int dz - a^2\int \frac{1}{z}dz$$ We can now integrate each part: $$z - a^2\ln|z| + C = (x^2 + a^2) - a^2\ln|x^2 + a^2| + C$$ Since this is our result from the substitution, we must add back the result from integration by parts: $$x\ln(x^2 + a^2) - 2((x^2 + a^2) - a^2\ln|x^2 + a^2| + C)$$

Simplifying and Writing the Final Answer

Finally, let's simplify the expression and write our final answer: $$x\ln(x^2 + a^2) - 2x^2 - 2a^2\ln|x^2 + a^2| + 4a^2 + C$$ The final answer is: $$\int \ln \left(x^{2}+a^{2}\right) d x = x\ln(x^2 + a^2) - 2x^2 - 2a^2\ln|x^2 + a^2| + 4a^2 + C$$

Key concepts

Substitution Method

The substitution method is a useful tool in calculus for evaluating integrals, particularly when integration by parts or standard rules don't easily apply. The main idea behind substitution is to simplify the integral by changing variables.To apply substitution:

• Identify a portion of the integrand that can be substituted with a new variable.
• Calculate the derivative of the new variable in terms of the old variable.
• Substitute all parts of the integral, including the differential, with expressions in terms of the new variable.
• Integrate with respect to the new variable.
• Reverse the substitution to express the integral back in the original variable.

In the given problem, we replaced the expression \(x^2 + a^2\) with \(z\), making it simpler to evaluate the integral. The substitution method can turn a complex integral into a more straightforward one, leading to easier solutions.

Definite and Indefinite Integrals

Integrals are classified as either definite or indefinite, depending on their properties and results.Indefinite integrals represent a family of functions and are often written as:\[\int f(x)dx = F(x) + C\]where \(F(x)\) is an antiderivative of \(f(x)\), and \(C\) is the arbitrary constant.Definite integrals, on the other hand, calculate the net area under a curve from one point to another and are expressed as:\[\int_{a}^{b} f(x)dx\]which yields a numeric value instead of a function.In our exercise, we dealt with an indefinite integral of \(\ln(x^2 + a^2)\) with respect to \(x\). The solution involves obtaining a general expression, incorporating an arbitrary constant \(C\), to represent the broad family of antiderivatives, showcasing the concept of indefinite integration.

Logarithmic Integration

Logarithmic integration is often required when an integrand includes a natural logarithm, such as \(\ln(x^2 + a^2)\) in the original exercise. These integrals can be challenging because they often require applying integration by parts or other techniques specifically suited to handle logarithmic functions.We used integration by parts here, choosing \(u = \ln(x^2 + a^2)\) to differentiate, since the derivative of a logarithmic function provides a simplifiable expression. This is crucial when dealing with logarithmic integration. By cleverly choosing the parts, we can reduce complexity, leading to a viable expression for the integral.Logarithmic functions have special properties and are frequently encountered in calculus, such as their characteristic slow growth and the fact that their derivative involves reciprocal functions. Understanding how to manage these through methods like integration by parts is vital for solving more elaborate calculus problems efficiently.