Question

Perform the indicated integrations. $$ \int \frac{d x}{9 x^{2}+18 x+10} $$

Short answer

The integral evaluates to \( \frac{1}{3} \tan^{-1}(3x + 3) + C. \)

Step-by-step solution

Identify the Form

The given integral is \( \int \frac{d x}{9 x^{2}+18 x+10} \). This integral is in the form of \( \int \frac{1}{ax^2+bx+c} \, dx \), which suggests the use of a proper substitution method or partial fraction decomposition if applicable.

Complete the Square

Completing the square for the quadratic expression in the denominator, \( 9x^2 + 18x + 10 \):\[ 9x^2 + 18x + 10 = 9(x^2 + 2x) + 10.\]Further, \( x^2 + 2x = (x+1)^2 - 1 \), so:\( 9(x^2 + 2x) = 9((x+1)^2 - 1) = 9(x+1)^2 - 9. \)Therefore, the expression becomes:\[ 9(x+1)^2 - 9 + 10 = 9(x+1)^2 + 1. \]

Rewrite the Integral

Substitute back into the integral:\[ \int \frac{dx}{9(x+1)^2 + 1}. \]This expression is favorable for a trigonometric substitution since it represents a form similar to a known inverse trigonometric function.

Use Trigonometric Substitution

Recognize the form \( a^2 + u^2 \) where \( u = x+1 \) and \( a = \frac{1}{3} \). Perform a trigonometric substitution: \( t = \frac{x+1}{1/3} \Rightarrow x + 1 = \frac{1}{3} t \Rightarrow dx = \frac{1}{3} dt \).Rewrite the integral:\[ \int \frac{\frac{1}{3}dt}{t^2 + 1}. \]

Identifying the Integral of a Trigonometric Form

The integral \( \int \frac{dt}{t^2 + 1} \) is a standard form that integrates to the inverse tangent function:\[ \frac{1}{3} \int \frac{dt}{t^2 + 1} = \frac{1}{3} \tan^{-1}(t) + C. \]Returning to the substitution \( t = 3(x + 1) \), this becomes:\[ \frac{1}{3} \tan^{-1}(3(x + 1)) + C. \]

Finalize the Solution

Substitute back \( t = 3x + 3 \) and finalize:\[ \frac{1}{3} \tan^{-1}(3x + 3) + C. \]Therefore, the integral evaluates to \( \frac{1}{3} \tan^{-1}(3x + 3) + C \) where \( C \) is the constant of integration.

Key concepts

Quadratic Expressions

A quadratic expression is a polynomial of degree two. It usually takes the form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. Quadratic expressions are encountered frequently in calculus, especially when dealing with integrals. Such expressions can often be simplified to a more workable form through methods like completing the square.

Completing the square is a technique used to transform a quadratic expression into a perfect square trinomial, making it easier to integrate or solve. For instance, if we start with the quadratic \( 9x^2 + 18x + 10 \), we can complete the square to rewrite it as \( 9(x+1)^2 + 1 \).

This transformation helps in revealing simpler forms, such as those related to trigonometric substitution. Recognizing these forms is key to applying appropriate integration techniques. It allows us to later integrate using known formulas, especially those involving trigonometric identities.

Trigonometric Substitution

Trigonometric substitution is an integration technique that involves substituting a trigonometric function for a variable. This method simplifies integrals involving square roots or expressions that can be expressed as sums or differences of squares. In the exercise, the expression \( 9(x+1)^2 + 1 \) was restructured into a form suitable for trigonometric substitution.

The trigonometric identities typically employed include:

• \( 1 - \sin^2\theta = \cos^2\theta \)
• \( 1 + \tan^2\theta = \sec^2\theta \)
• \( \sinh^2\theta + 1 = \cosh^2\theta \)

For this particular problem, the identity \( 1 + \tan^2\theta = \sec^2\theta \) was utilized.

By letting \( t = \frac{x+1}{1/3} \), we transformed the expression \( x+1 \) into \( \frac{1}{3} t \), which simplified the integral into a standard trigonometric form \( \int \frac{dt}{t^2 + 1} \). The objective is to align your integral with well-known integral formulas, making the solution process much less complex.

Inverse Trigonometric Functions

Inverse trigonometric functions are essential in calculus for integrating certain types of expressions. They reverse the process of trigonometric functions. Common inverse trigonometric functions include \( \sin^{-1}, \cos^{-1}, \tan^{-1} \), and their effectiveness becomes apparent when the integral corresponds to a derivative of an inverse trig function.

In this case, the integral \( \int \frac{dt}{t^2 + 1} \) was solved using \( \tan^{-1}(t) \). The standard formula \( \int \frac{1}{t^2 + 1} \, dt = \tan^{-1}(t) + C \) was employed, but scaled by a factor derived during substitution.

Becoming familiar with inverse functions is vital because they often appear in solutions to integrals involving quadratic expressions transformed by trigonometric substitutions. These functions simplify the integration process and provide straightforward results, such as \( \frac{1}{3} \tan^{-1}(3x + 3) + C \), which finalizes the integral evaluation in a manageable form.