Question

Multiple substitutions If necessary, use two or more substitutions to find the following integrals. $$\int \frac{dx}{\sqrt{1+\sqrt{1+x}}}(\text{ Hint: Begin with }u=\sqrt{1+x}\text{ ) }$$

Short answer

Question: Evaluate the integral: $$\int \frac{1}{\sqrt{1+\sqrt{1+x}}}dx$$ Answer: $$4\sqrt{(x)((1+\sqrt{1+x})}+C$$

Step-by-step solution

First Substitution

We are given the hint to start with the substitution $u=\sqrt{1+x}$. Let's find the differential $du$ as well: $$du=\frac{1}{2\sqrt{1+x}}dx$$ We can rearrange the equation to isolate $dx$: $$dx=2\sqrt{1+x}du$$ Now we substitute $u$ and $dx$ into the integral: $$\int \frac{1}{\sqrt{1+\sqrt{1+x}}}dx=\int \frac{1}{\sqrt{1+u}}2\sqrt{1+x}du$$

Second Substitution

To simplify the integral further, we can make another substitution. Let's let $v=1+u$, so $dv=du$. Now we can rewrite the integral: $$\int \frac{2\sqrt{1+x}}{\sqrt{1+u}}du=\int \frac{2\sqrt{1+x}}{\sqrt{v}}dv$$ Notice that $1+x=u^2-1$. So the integral becomes: $$\int \frac{2\sqrt{u^2-1}}{\sqrt{v}}dv$$

Integration

Now we can integrate the expression with respect to $v$: $$\int \frac{2\sqrt{u^2-1}}{\sqrt{v}}dv=2\int \frac{\sqrt{u^2-1}}{\sqrt{v}}dv$$ This is the integral of a function in the form $\frac{u}{\sqrt{v}}$, where $u=\sqrt{u^2-1}$ and $v=v$. Integrating with respect to $v$, we get: $$2\int \frac{\sqrt{u^2-1}}{\sqrt{v}}dv=4\sqrt{u^2-1}\sqrt{v}+C$$

Back-substitution

Now we substitute the original variables back into the expression. Recall that $v=1+u$, and $u=\sqrt{1+x}$: $$4\sqrt{(u^2-1)(1+u)}+C=4\sqrt{({\sqrt{1+x}}^2-1)((1+\sqrt{1+x})}+C$$ Simplifying, we get the final answer: $$4\sqrt{(x)((1+\sqrt{1+x})}+C$$

Key concepts

U-Substitution

The u-substitution is a powerful tool for solving integrals, which is essentially the reverse process of the chain rule in differentiation. It allows us to transform complex integrals into simpler forms that are easier to evaluate. To apply u-substitution effectively, we select a part of the integral to be our new variable u, which often contains the inner function of a composition.

As in our example, we begin by choosing u = $\sqrt{1+x}$, then differentiate u to find du, and express dx in terms of du. This replaces the variable x with u, simplifying the integral. One crucial step often involves expressing all instances of the original variable x using the new variable u before integrating. Sometimes, multiple substitutions, as in this example, are necessary to unravel the integral into a form that can be integrated.

Integration Techniques

Apart from u-substitution, calculus students should be familiar with a variety of integration techniques to solve indefinite integrals. Common methods include integration by parts, trigonometric substitution, partial fractions, and integration of rational functions using polynomial long division. Each technique has its own set of rules and suitable scenarios.

For example, integration by parts is often used when an integral is a product of two functions. Trigonometric substitution is helpful when dealing with square roots involving squares of variables. Understanding when and how to apply these methods can make a significant difference in solving complex integrals efficiently. The key is often to recognize patterns and select the appropriate technique for the task at hand.

Indefinite Integrals

An indefinite integral, represented by $\int f(x)dx$, is essentially the collection of all antiderivatives of a given function f(x). Unlike definite integrals, indefinite integrals do not have limits of integration and therefore include an arbitrary constant, C, representing the family of all possible antiderivatives.

To solve an indefinite integral, we need to find a function whose derivative is the integrand. This process can range from simple to quite intricate, as some integrals might require inventive substitutions or manipulations to find a solvable form. Always ensure that the final answer is simplified and replace any substitution variables back to the original variable.

Calculus

Calculus is a branch of mathematics focused on understanding changes. It is divided mainly into two areas: differential calculus, concerning rates of change and slopes of curves, and integral calculus, concerning accumulation of quantities and areas under and between curves.

Integration, a key component of integral calculus, is about putting together the parts to find the whole. While differentiation breaks apart functions to find the derivative (rate of change), integration takes a function apart to find the area under the curve (the integral). When faced with a difficult integration problem, understanding the fundamental principles of calculus and having a repertoire of techniques at your disposal is essential. The process of learning calculus involves practice, intuition, and the ability to link different concepts for problem-solving.