Problem 87 - Calculus for Biology and Medicine

Question

Compute the indefinite integrals. $$ \int \frac{2 x^{2}}{x^{2}+1} d x $$

Step-by-Step Solution

Answer

The indefinite integral is \( (x^2 + 1) - \ln |x^2 + 1| + C \).

Step 1: Apply Substitution

To solve the integral \( \int \frac{2x^2}{x^2+1} \, dx \), we use the substitution method by letting \( u = x^2 + 1 \). Then, differentiate \( u \) with respect to \( x \) to find \( \frac{du}{dx} = 2x \), or equivalently, \( du = 2x \, dx \). We notice that we need \( 2x^2 \, dx \), which can be rewritten using \( u - 1 = x^2 \) to substitute for \( x^2 \). Hence, the integral becomes \( \int \frac{u - 1}{u} \, du \).

Step 2: Simplify the Integral

Now that we have \( \int \frac{u-1}{u} \, du \), we can separate into two different integrals as follows: \( \int (1 - \frac{1}{u}) \, du \). This separation is valid because \( \frac{u-1}{u} = 1 - \frac{1}{u} \).

Step 3: Integrate Each Term

The integral \( \int 1 \, du \) is simply \( u \), and the integral \( - \int \frac{1}{u} \, du \) is \( -\ln |u| \). Therefore, the solution becomes \( u - \ln |u| + C \), where \( C \) is the constant of integration.

Step 4: Substitute Back for \( u \)

Re-substitute \( u = x^2 + 1 \) back into the expression obtained from integration. Therefore, the integral becomes \( (x^2 + 1) - \ln |x^2 + 1| + C \).

Key Concepts

Integration by Substitution

Integration by substitution is a technique often used when faced with complex integrals. This approach involves replacing a part of the integral with a new variable, making it easier to compute. Think of it as transforming the integral into a simpler form.
In our exercise, we encounter the integral \( \int \frac{2x^2}{x^2+1} \, dx \). This integral is tricky, but substitution can simplify it. We choose \( u = x^2 + 1 \), reflecting one part of the expression under integration. This replacement helps isolate the variable \( x \), allowing us to manage the integral more comfortably.
The derivative of \( u \), \( \frac{du}{dx} = 2x \), leads us to \( du = 2x \, dx \). By substituting these values into the integral, we change its form, which simplifies the process. The integral transforms into \( \int \frac{u - 1}{u} \, du \), leading us smoothly to the next steps.

Algebraic Manipulation

Once we've employed substitution, we often need algebraic manipulation to further simplify an integral. This involves breaking or combining expressions to make integration manageable.
In our solution, \( \int \frac{u - 1}{u} \, du \) splits into two more convenient integrals: \( \int (1 - \frac{1}{u}) \, du \). Algebraically, \( \frac{u-1}{u} \) simplifies to \( 1 - \frac{1}{u} \). This separation is critical, as it allows us to handle the integral in parts, integrating terms individually.
Handling each term becomes straightforward: \( \int 1 \, du \) and \( - \int \frac{1}{u} \, du \). These simplified forms are familiar and easier to integrate, a direct result of effective algebraic manipulation.

Constant of Integration

In indefinite integrals, the constant of integration, denoted as \( C \), is an essential element. It represents the family of solutions arising from indefinite integrating an expression.
When we evaluate each part of our split integral, \( \int 1 \, du \) results in \( u \) and \( - \int \frac{1}{u} \, du \) results in \( - \ln |u| \). Combining these, we get \( u - \ln |u| + C \).
The constant \( C \) ensures that we account for any vertical shifts in the integral's graph, which is crucial since indefinite integrals include all possible antiderivatives of a function. Thus, \( C \) represents the infinite nature of these possible solutions. Always remember to add \( C \) when presenting your final answer in indefinite integration.