Question

Evaluate the following integrals: $$ \int \frac{6 x-5}{\sqrt{3 x^{2}-5 x+1}} d x $$

Short answer

Question: Determine the value of the following integral: $$ \int \frac{6x - 5}{\sqrt{3x^2 - 5x + 1}} dx $$ Solution: The value of the integral is given by: $$ \int \frac{6x - 5}{\sqrt{3x^2 - 5x + 1}} dx = 2(3x^2 - 5x + 1)^{\frac{1}{2}} + C $$ where \(C\) is the constant of integration.

Step-by-step solution

Identify the appropriate substitution

For the given integral, choose a substitution so that the derivative of a function in the numerator matches with a function in the denominator. In this case, let's choose: $$ u = 3x^2 - 5x + 1 $$ Next, find the derivative of \(u\) concerning \(x\): $$ \frac{d u}{d x} = 6x - 5 $$ It matches with the numerator of the integrand, so this substitution is suitable.

Perform the substitution

Now that we have our substitution, let's rewrite the integral in terms of \(u\). Replace \(6x - 5\) with \(du/dx\) and \(\sqrt{3x^2 - 5x + 1}\) with \(\sqrt{u}\), and substitute \(dx\): $$ \int \frac{6x - 5}{\sqrt{3x^2 - 5x + 1}} dx = \int \frac{1}{\sqrt{u}} \frac{d u}{d x} d x $$ Now, cancel out \(dx\) from both numerator and denominator: $$ \int \frac{1}{\sqrt{u}} \frac{d u}{d x} d x = \int \frac{1}{\sqrt{u}} d u $$ We have now simplified the integral in terms of \(u\).

Integrate with respect to \(u\)

Now, we can proceed to integrate the simplified expression with respect to \(u\): $$ \int \frac{1}{\sqrt{u}} d u $$ We can rewrite the integrand as \(u^{-1/2}\): $$ \int u^{-\frac{1}{2}} d u $$ Now, we may integrate: $$ \int u^{-\frac{1}{2}} d u = \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C = 2u^{\frac{1}{2}} + C $$ Here, \(C\) is the constant of integration.

Substitute back for \(x\)

Now that we have integrated with respect to \(u\), let's substitute back for \(x\) using \(u = 3x^2 - 5x + 1\): $$ 2u^{\frac{1}{2}} + C = 2(3x^2 - 5x + 1)^{\frac{1}{2}} + C $$ So the final solution is: $$ \int \frac{6x - 5}{\sqrt{3x^2 - 5x + 1}} dx = 2(3x^2 - 5x + 1)^{\frac{1}{2}} + C $$

Key concepts

Substitution Method

The substitution method in integral calculus is akin to a magical transformation, simplifying complex integrals into more manageable forms. It involves introducing a new variable, typically denoted as u, which represents a function within the integrand. By differentiating this function with respect to x to obtain du/dx, we gain the power to replace parts of the original integral.

Think of it as putting on a pair of glasses that helps you see through the clutter of an equation. In our problem, we chose u = 3x^2 - 5x + 1 because its derivative, 6x - 5, matches the numerator of our integrand. This strategic choice is pivotal, transforming the integral into a function of u that we can integrate more directly. The ultimate goal here is to rewrite the integral in such a way that the integration process becomes clear-cut and straightforward.

Indefinite Integration

Indefinite integration, often considered the antithesis of differentiation, seeks to find the original function from its rate of change - much like retracing your steps back home. Integral signs, long s-looking symbols, are guideposts indicating that the process of integration is at play.

In our mathematical journey, indefinite integration does not provide a fixed destination—the 'constant of integration', denoted by C, hints at an array of potential functions, all differing by a constant value. After applying the substitution method, the next phase involves integrating with respect to u to find the function that, when differentiated, would yield our integrand. To accomplish this, we employ anti-differentiation techniques, effectively reversing the gears of differentiation. In the context of our example, we integrated u^{-1/2} to obtain 2u^{1/2} + C.

Integrals of Rational Functions

Rational functions are fractions where both the numerator and denominator are polynomials. The intrigue of integrating such fractions lies in their complexity — they often demand creative approaches like the substitution method to uncover an antiderivative.

In the case of our problem, we are working with a rational function where the denominator is under a square root, further complicating matters. However, fear not, as the path to resolution becomes clear once a suitable substitution is found. By examining the structure of the rational function, identifying the component whose derivative appears elsewhere in the integral, and performing the substitution, we transform the rational function into a simpler expression that we can integrate with confidence. The resulting function often reveals insights into the behavior of the original function that were not immediately obvious.