Question

Evaluate the following integrals : $$\int \frac{d x}{x \sqrt{\left(x^{2}-x+2\right)}}$$

Short answer

Question: Evaluate the integral \(\int \frac{d x}{x \sqrt{\left(x^{2}-x+2\right)}}\). Answer: \(\int \frac{d x}{x \sqrt{\left(x^{2}-x+2\right)}} = \ln|x| + C\)

Step-by-step solution

Recognize the form of the expression under the square root

The expression under the square root is \(x^2-x+2\). This expression can be rewritten as \(x^2-x+1+1 = (x-\frac{1}{2})^2 + 1\), which resembles the expression \(a^2 + x^2\), which is suitable for trigonometric substitution.

Choose a suitable trigonometric substitution

Since the expression under the square root resembles \(a^2 + x^2\), the substitution \(x-\frac{1}{2} = a\tan\theta\) is appropriate. Here, \(a=1\), so the substitution becomes \(x-\frac{1}{2} = \tan\theta\). Therefore, \(x = \tan\theta + \frac{1}{2}\).

Find the differential of x

With \(x = \tan\theta + \frac{1}{2}\), differentiate \(x\) with respect to \(\theta\) to get: \(\frac{dx}{d\theta} = \sec^2\theta\). Thus, \(dx = \sec^2\theta d\theta\).

Substitute the trigonometric expression and the differential into the integral

Substitute \(x-\frac{1}{2} = \tan\theta\) and \(dx = \sec^2\theta d\theta\) into the integral: $$\int \frac{d x}{x \sqrt{\left(x^{2}-x+2\right)}} = \int \frac{\sec^2\theta d\theta}{(\tan\theta + \frac{1}{2}) \sqrt{1+\tan^2\theta}}$$

Simplify the integral

Now, simplify the integral with the trigonometric identity \(\sec^2\theta=1+\tan^2\theta\): $$\int \frac{\sec^2\theta d\theta}{(\tan\theta + \frac{1}{2}) \sqrt{1+\tan^2\theta}} = \int \frac{d\theta}{\tan\theta + \frac{1}{2}}$$

Find the antiderivative

Let \(u=\tan\theta+\frac{1}{2}\), then \(du=d\theta\). The integral becomes: $$\int \frac{d\theta}{\tan\theta + \frac{1}{2}} = \int \frac{du}{u}$$ Now integrate with respect to \(u\): $$\int \frac{du}{u} = \ln|u| + C$$

Apply the inverse substitution

Now, substitute back \(u=\tan\theta+\frac{1}{2}\): $$\ln|u| + C = \ln|\tan\theta + \frac{1}{2}| + C$$ Next, substitute back \(x-\frac{1}{2} = \tan\theta\): $$\ln|\tan\theta + \frac{1}{2}| + C = \ln|x-\frac{1}{2} + \frac{1}{2}| + C$$ The final answer is: $$\int \frac{d x}{x \sqrt{\left(x^{2}-x+2\right)}} = \ln|x| + C$$

Key concepts

Trigonometric Substitution

In integral calculus, especially when dealing with integrals involving square roots, trigonometric substitution is a technique that can simplify the integration process. The idea behind this method is to substitute a trigonometric function for another expression to take advantage of trigonometric identities.

For instance, when you encounter an expression like \(a^2 + x^2\), trigonometric substitution suggests that substituting \(x = a\tan(\theta)\) will be useful, because the identity \(1 + \tan^2(\theta) = \sec^2(\theta)\) allows for simplification. Similarly, for expressions like \(a^2 - x^2\) or \(x^2 - a^2\), substitutions involving \(\sin(\theta)\) or \(\sec(\theta)\) would be appropriate.

To apply trigonometric substitution:

• Identify an expression that fits the form suitable for substitution.
• Select the right trigonometric function to substitute.
• Find the differential of the new trigonometric expression.
• Substitute into the integral and simplify using trigonometric identities.
• Perform the integration.
• Substitute back trigonometric terms to the original variable.

Using this method can turn a challenging integral into one that's more straightforward to solve.

Antiderivatives

Antiderivatives are the reverse process of differentiation. In other words, finding an antiderivative means looking for a function whose derivative is the given function. This is at the heart of integral calculus; the antiderivative of a function represents the indefinite integral of that function.

In practice, finding an antiderivative involves recognizing patterns and using known derivatives. For instance, the antiderivative of \(\frac{1}{x}\) is \(\ln|x|\) because the derivative of \(\ln|x|\) is \(\frac{1}{x}\). When performing integration, you look for a function that would differentiate to yield the integrand (the function you are integrating). Often, this process requires manipulation of the integrand using algebraic techniques or substitutions to match a known derivative.

Fundamentally, each function (within certain classes) has an infinite number of antiderivatives, but these differ only by a constant term. The arbitrary constant, typically denoted \(C\), is crucial because it accounts for the ‘+ constant’ that appears when differentiating a function. Consequently, the general solution to an indefinite integral is expressed as the antiderivative plus \(C\).

Definite and Indefinite Integrals

Integral calculus is divided into two fundamental types of integrals: definite and indefinite. An indefinite integral is represented by the antiderivative of a function, as it does not have upper or lower limits. It includes the constant \(C\) because without specific limits, there's an infinite number of possible values.

On the flip side, a definite integral has clear limits, usually denoted as \(a\) and \(b\), representing the lower and upper bounds of the integration, respectively. The definite integral calculates the area under the curve defined by the function over the interval from \(a\) to \(b\). It is expressed as \(\int_a^b f(x)\,dx\), and it produces a specific numerical value, rather than a function with a '+ \(C\)'.

To solve a definite integral, one typically finds the indefinite integral (antiderivative) first and then applies the Fundamental Theorem of Calculus, which involves subtracting the antiderivative evaluated at \(a\) from the antiderivative evaluated at \(b\). This process is often described as 'evaluating the antiderivative at the bounds' and leads to the actual area or accumulated quantity that the definite integral signifies.