Problem 61 - Calculus Single Variable

Question

Calculate the given integral. $$ \int \frac{x}{\sqrt{x^{2}+2 x+2}} d x $$

Step-by-Step Solution

Answer

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

Step 1: Recognize substitution

Notice that the expression under the square root, \(x^2 + 2x + 2\), suggests a trigonometric substitution or completing the square. Complete the square: \(x^2 + 2x + 2 = (x + 1)^2 + 1\). We can use this adjusted form for substitution.

Step 2: Use trigonometric substitution

Substitute \(x + 1 = \tan \theta\), which implies \(dx = \sec^2 \theta \, d\theta\). The square root becomes \(\sqrt{(x + 1)^2 + 1} = \sqrt{\tan^2 \theta + 1} = \sec \theta\). Substitute these into the integral.

Step 3: Simplify the integral

Substitute into the integral to get: \[ \int \frac{\tan\theta - 1}{\sec\theta} \cdot \sec^2 \theta \, d\theta = \int (\tan \theta \cdot \sec \theta - \sec \theta) \, d\theta. \] This simplifies to \(\int \sec\theta \tan\theta \, d\theta - \int \sec\theta \, d\theta\).

Step 4: Integrate using known derivatives

The integral \(\int \sec\theta \tan\theta \, d\theta\) is \(\sec \theta\). The integral \(\int \sec\theta \, d\theta\) is \(\ln |\sec\theta + \tan\theta| + C\). Hence, the overall integral becomes \(\sec\theta - \ln |\sec\theta + \tan\theta| + C\).

Step 5: Back-substitute using original variable

Since \(x + 1 = \tan\theta\), we have \(\sec\theta = \sqrt{1 + \tan^2 \theta} = \sqrt{1 + (x+1)^2} = \sqrt{x^2+2x+2}\). Then \(\tan\theta = x + 1\). Substitute back in terms of \(x\): \[ \sqrt{x^2+2x+2} - \ln |\sqrt{x^2+2x+2} + (x+1)| + C. \]

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique in calculus for evaluating integrals, especially when they involve square roots of quadratic expressions. In this method, we replace the original variable with a trigonometric function. This often transforms the integrand into a more easily integrable form. In the example provided, the function under the square root is completed into a perfect square, giving us \((x+1)^2 + 1\). Seeing this, we choose to make the substitution \(x + 1 = \tan \theta\).

• This substitution implies \( dx = \sec^2 \theta \, d\theta \).
• With this, \( \sqrt{(x+1)^2 + 1} = \sec \theta \).

The integral simplifies significantly because the trigonometric identities allow for straightforward integration. The choice of substitution depends on recognizing forms similar to \(1 + \tan^2 \theta = \sec^2 \theta\). Trigonometric substitution is ideal when you see expressions such as \(a^2 - x^2\), \(a^2 + x^2\), or \(x^2 \/ a^2 - 1\), where converting \(x\) into trig functions can simplify the expression for easier integration.

Completing the Square

Completing the square is a fundamental algebraic technique that facilitates many calculus problems, particularly when dealing with quadratic expressions. The process involves rewriting a quadratic expression in the form \(ax^2 + bx + c\) as a perfect square plus or minus some constant. For example, for the expression \(x^2 + 2x + 2\), we rewrite it as \((x+1)^2 + 1\).

• Start by identifying the linear coefficient and divide it by 2, then square the result to form a perfect square trinomial.
• Here, take \(1\) (half of \(2\)) and square it to get \(1\), leading to \((x+1)^2\).

Completing the square is especially useful because it allows us to directly recognize \(\tan \theta\) or \(\sin \theta\) substitutions for integrals by transforming the expression neatly.This method is critical when working with quadratic expressions under a square root as it paves the way for trigonometric substitution, simplifying the integral process.

Definite Integrals

Definite integrals compute the exact area under a curve described by a function between two specified points. Unlike indefinite integrals that yield a family of functions with an arbitrary constant, definite integrals result in a single numerical value.Calculating a definite integral involves evaluating the antiderivative at the upper limit and lower limit, then subtracting these values. For example, we might integrate \(\int_{a}^{b} f(x) \, dx\) to find the total accumulation of the quantity described by \(f(x)\) between \(x=a\) and \(x=b\).

• The formula for evaluating is \[ F(b) - F(a) \]
• Here, \(F(x)\) represents the antiderivative of \(f(x)\).

In practical applications, definite integrals are used to calculate quantities like area under a curve, volumetric changes, and even some cases of physical displacement. Understanding definite integrals is essential for applying calculus to real-world problems and analysis.