Question

Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals. \(\int \frac{d x}{1+\sqrt{x+1}}\)

Short answer

\[2 ( \sqrt{x+1} - \ln |1+\sqrt{x+1}| ) + C\]

Step-by-step solution

Define a Substitution

To simplify the integral, we will start by defining a substitution to remove the square root from the denominator. Let \( u = \sqrt{x + 1} \), so \( u^2 = x + 1 \) which means \( x = u^2 - 1 \). The derivative \( \frac{dx}{du} = 2u \) leads to \( dx = 2u \, du \).

Substitution in the Integral

Substitute \( u \) and \( dx \) in terms of \( u \) into the integral. The integral becomes:\[\int \frac{dx}{1+\sqrt{x+1}} = \int \frac{2u}{1+u} \, du = 2 \int \frac{u}{1+u} \, du\]

Simplify the Integrand

Simplify \( \frac{u}{1+u} \) by rewriting it as \( \frac{u+1-1}{1+u} = 1 - \frac{1}{1+u} \). So the integral becomes:\[2 \int \left( 1 - \frac{1}{1+u} \right) \, du = 2 \left( \int 1 \, du - \int \frac{1}{1+u} \, du \right)\]

Integrate Each Term Separately

Evaluate each integral separately:1. The integral of constant 1 is \( \int 1 \, du = u \).2. The integral \( \int \frac{1}{1+u} \, du = \ln |1+u| + C_1 \).Thus, the combined integral is:\[2 \left( u - \ln |1+u| \right) + C\]

Back Substitute to Original Variable

Substitute back to \( x \) by replacing \( u \) with \( \sqrt{x+1} \):\[2 \left( \sqrt{x+1} - \ln |1+\sqrt{x+1}| \right) + C\]

Final Result

Thus, the evaluated integral is:\[\int \frac{d x}{1+\sqrt{x+1}} = 2 \left( \sqrt{x+1} - \ln |1+\sqrt{x+1}| \right) + C\]

Key concepts

Substitution Method

The substitution method is a fundamental tool in calculus used to simplify complex integrals. At its core, substitution involves replacing a part of the integral that appears challenging, often transforming it into a more manageable form. This is especially useful when dealing with functions that involve complicated expressions, like square roots or trigonometric functions.

In our example, we used the substitution \( u = \sqrt{x+1} \) to simplify the integral. This choice was strategic because it directly helped eliminate the square root in the denominator, leading to a simpler form \( \int \frac{2u}{1+u} \, du \).

When using substitution, always remember these key steps:

• Select an appropriate substitution that simplifies the problem.
• Calculate the derivative of your substitution \( u \) with respect to \( x \), which helps in changing \( dx \) in terms of \( du \).
• Rewrite the integral in terms of the new variable \( u \).
• Don't forget to back-substitute the original variable once the integration is complete!

By following these steps, substitution can make otherwise daunting integrals much easier to handle.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions that are easier to integrate. This method proves particularly helpful when dealing with rational functions, where the degree of the polynomial in the denominator is higher.

In this exercise, we applied partial fraction decomposition after simplifying the integrand to \( \frac{u}{1+u} \). This expression could be rewritten as \( 1 - \frac{1}{1+u} \), separating it into two much simpler integrals: \( \int 1 \, du \) and \( \int \frac{1}{1+u} \, du \).

Here's a simple checklist for doing partial fraction decomposition:

• Ensure that the numerator's degree is less than the denominator's; if not, perform polynomial division first.
• Factor the denominator into its irreducible factors if possible.
• Express the rational function as a sum of simpler fractions, assigning each factor a simple term in the decomposition.
• Solve for the coefficients by equating the original expression to the sum of your proposed fractions.
• Once decomposed, integrate each simple fraction separately.

This method allows you to tackle complex rational functions piece by piece, making integration more manageable.

Rational Functions

Rational functions are expressions that represent the quotient of two polynomials. In calculus, these functions are commonly encountered when integrating expressions, especially those requiring advanced techniques like partial fractions or substitution.

Understanding rational functions is key because:

• They often appear in various forms that require simplification before integration.
• Identification of rational functions can help choose the appropriate method for integration, such as partial fraction decomposition.
• Simplifying a rational function can make integration doable, often leading to results that are expressed as elementary functions.

In our case, we dealt with the expression \( \frac{dx}{1+\sqrt{x+1}} \), which turned into a rational function after substitution. Simplifying it into \( \frac{u}{1+u} \) made integration possible using partial fraction decomposition.

Mastering the properties and characteristics of rational functions not only aids in integration but also prepares you for tackling more complex calculus problems.

Definite Integrals

While the exercise at hand dealt with an indefinite integral, definite integrals form an essential part of calculus and provide more comprehensive applications, such as calculating areas or solving real-world problems.

Definite integrals differ primarily because they evaluate the signed area under a curve between two limits, say \( a \) and \( b \). The formula is given by:\[\int_a^b f(x) \, dx = F(b) - F(a)\]where \( F(x) \) is an antiderivative of \( f(x) \).

Key aspects to understand with definite integrals include:

• They provide a numeric result representing an area, unlike indefinite integrals that include a constant of integration \( C \).
• Adjustments during integration must account for these limits, often requiring changes in variables similar to our substitution method.
• Understanding definite integrals enriches comprehension of how calculus applies to physics, engineering, and other fields.

When practicing with integrals, consider transitioning from indefinite to definite integrals by applying limits, giving real-world context to mathematical operations.