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Use the Table of Integrals to compute each integral after manipulating the integrand in a suitable way. $$ \int \sqrt{x^{2}+2 x+2} d x $$

Short Answer

Expert verified
\( \frac{1}{2}\left(\sqrt{x^2+2x+2}\cdot (x+1) + \ln\left|\sqrt{x^2+2x+2} + (x+1)\right|\right) + C \)

Step by step solution

01

Complete the Square

Notice that the expression inside the square root is a quadratic expression, specifically \(x^2 + 2x + 2\). To simplify the integration, first complete the square for the quadratic expression. This is done by rewriting it in the form \((x + a)^2 + b\).\[x^2 + 2x + 2 = (x + 1)^2 + 1\]This is because \((x + 1)^2 = x^2 + 2x + 1\). Thus we must add another 1 to complete the original expression.
02

Apply a Trigonometric Substitution

With the form \((x + 1)^2 + 1\), you can simplify the integral by using the trigonometric substitution \(x + 1 = \tan(\theta)\). This means that \(dx = \sec^2(\theta)\, d\theta\) and the integrand becomes \(\sqrt{\tan^2(\theta) + 1} = \sec(\theta)\). Substitute into the integral: \[\int \sec(\theta) \cdot \sec^2(\theta) \, d\theta = \int \sec^3(\theta) \, d\theta\]
03

Integrate Using the Secant Cubed Integral Formula

Using a table of integrals, you can find the integral of \(\sec^3(\theta)\), which is a standard integral in calculus:\[\int \sec^3(\theta) \, d\theta = \frac{1}{2}(\sec(\theta)\tan(\theta) + \ln|\sec(\theta) + \tan(\theta)|) + C\]
04

Substitute Back to Variable \(x\)

Now we need to substitute back in terms of \(x\). Recall the substitution \(x+1 = \tan(\theta)\), which implies \(\theta = \tan^{-1}(x+1)\). Therefore, \(\sec(\theta) = \sqrt{(x+1)^2 + 1}\) and \(\tan(\theta) = x+1\). Substitute these back into the solution:\[\frac{1}{2}\left(\sqrt{x^2+2x+2}\cdot (x+1) + \ln\left|\sqrt{x^2+2x+2} + (x+1)\right|\right) + C\]
05

Final Step: Simplified Result

Therefore, the integral \(\int \sqrt{x^2 + 2x + 2} \, dx\) is:\[\frac{1}{2}\left(\sqrt{x^2+2x+2}\cdot (x+1) + \ln\left|\sqrt{x^2+2x+2} + (x+1)\right|\right) + C\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Substitution
Trigonometric substitution is a powerful technique in calculus used to simplify integrals, particularly those involving square roots of quadratic expressions. It converts algebraic expressions involving square roots into trigonometric functions, which can be easier to integrate.

This method usually involves substituting a trigonometric function for a variable to simplify the square root. For instance, in the integral \(\int \sqrt{(x + 1)^2 + 1} \, dx\), we apply the substitution \(x + 1 = \tan(\theta)\).

This choice is strategic because \(\tan^2(\theta) + 1 = \sec^2(\theta)\), making the expression inside the square root simply \(\sec(\theta)\). Here’s how it unfolds:
  • Derive \(dx\) from the substitution: \(dx = \sec^2(\theta) \, d\theta\).
  • Replace the square root and \(dx\) in the integral: \(\int \sec(\theta) \cdot \sec^2(\theta) \, d\theta = \int \sec^3(\theta) \, d\theta\).
  • This transforms the original expression into a format that aligns with standard integrals.
Completing the Square
Completing the square is a technique mostly used to simplify quadratic expressions, turning them into a perfect square plus or minus a constant. This simplification makes it easier to handle integrals and derivatives of these expressions.

The process is straightforward and involves a few steps:
  • Take the quadratic expression, in this case, \(x^2 + 2x + 2\).
  • Identify terms to form a perfect square: \((x + a)^2\).
  • Rewrite the expression: \((x + 1)^2 + 1\), since \((x + 1)^2 = x^2 + 2x + 1\).

By completing the square, you transform the integrand into a format amenable to trigonometric substitution. This method is highly useful in a variety of calculus problems, helping to simplify expressions for better integration and differentiation.
Standard Integrals
Standard integrals refer to a set of commonly used integral formulas that students and professionals memorize for convenience. These are crucial building blocks in calculus, providing quick solutions to basic integrals.

In the exercise, we simplify and convert our integral through trigonometric substitution to align it with a standard integral. Specifically, the exercise involves the integral of \(\sec^3(\theta)\):

\[\int \sec^3(\theta) \, d\theta = \frac{1}{2}(\sec(\theta)\tan(\theta) + \ln|\sec(\theta) + \tan(\theta)|) + C\]

Memorizing this integral, among others, helps in simplifying and solving complex integrals without the need to perform the integration from scratch every time.
  • They serve as powerful tools when tackling various calculus problems.
  • Using standard integrals allows students to focus on restructuring problems to fit known formats.

By leveraging these formulas, complex integrals become manageable and straightforward.

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Most popular questions from this chapter

Compute the Taylor polynomial of degree \(n\) about \(x=0\) for each function and compare the value of the function at the indicated point with the value of the corresponding Taylor polynomial. $$ f(x)=\sqrt{1+x}, n=3, x=0.1 $$

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Use the Table of Integrals to compute each integral after manipulating the integrand in a suitable way. $$ \int_{1}^{e}(x+2)^{2} \ln x d x $$

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