Question

Evaluate the following integrals: $$ \int \frac{\left(2 x^{2}-3 x\right) d x}{\sqrt{x^{2}-2 x+5}} $$

Short answer

Question: Evaluate the integral \(\int \frac{\left(2 x^{2}-3 x\right) d x}{\sqrt{x^{2}-2 x+5}}\). Answer: \(\left[\sqrt{x^2 - 2x + 5}\right] + C\)

Step-by-step solution

Observe the expression inside the square root

First, let's look at the expression inside the square root: \(x^2 - 2x + 5\). Now, we can differentiate this with respect to \(x\) to see if we can get the expression in the numerator.

Differentiate the expression inside the square root

Differentiating \(x^2 - 2x + 5\) with respect to \(x\) gives us: $$ \frac{d}{dx}(x^2 - 2x + 5) = 2x - 2 $$

Perform substitution to simplify the integral

Notice that the result of the differentiation is close to the expression in the numerator, but not exactly the same. However, we can introduce a constant while differentiating to get the actual expression. Let's use the substitution: $$ u = x^2 - 2x + 5 $$ Differentiating both sides with respect to \(x\) and multiplying by a constant \(k\), we get: $$ k \frac{du}{dx} = k(2x - 2) $$ Now, we can choose \(k = 2\) to obtain the expression in the numerator as: $$ 2\left(2x-2\right) = 4x - 4 = 2x^2 - 3x $$ Now, let's manipulate the original integral:

Substitute the variables in the integral

Manipulating the integral with the result of the substitution, we get: $$ \int \frac{\left(2 x^{2}-3 x\right) d x}{\sqrt{x^{2}-2 x+5}} = \int \frac{\frac{1}{2}\left(4x-4\right) dx}{\sqrt{u}} $$ From the substitution, we have \(\frac{du}{dx} = 2(2x - 2)\). So, we can write \(dx = \frac{du}{4}\), and substitute it in the integral: $$ = \int \frac{\frac{1}{2}\left(4x-4\right) \frac{du}{4}}{\sqrt{u}} $$ Now, the integral in terms of \(u\) becomes: $$ = \frac{1}{8} \int \frac{4x - 4}{\sqrt{u}} du $$

Solve the integral with respect to u

Now we can solve the integral in terms of \(u\): $$ = \frac{1}{8} \int \frac{4x - 4}{\sqrt{u}} du = \frac{4}{8} \int \frac{1}{\sqrt{u}} du = \frac{1}{2} \int u^{-\frac{1}{2}} du $$ Integrating with respect to \(u\): $$ = \frac{1}{2} \left[2u^{\frac{1}{2}}\right] + C $$

Substitute back to the original variable

Now, let's substitute back our original variable, \(x\), by replacing \(u = x^2 - 2x + 5\): $$ = \left[\sqrt{x^2 - 2x + 5}\right] + C $$ So, the final answer for the integral is: $$ \int \frac{\left(2 x^{2}-3 x\right) d x}{\sqrt{x^{2}-2 x+5}} = \left[\sqrt{x^2 - 2x + 5}\right] + C $$

Key concepts

Integration by Substitution

Integration by substitution, often referred to as u-substitution, is a technique used in integral calculus to simplify complex integrals. The method involves changing the variable of integration to something simpler, which in turn, transforms the integral into an easier form. To effectively employ this method, one needs to identify a piece of the integrand that, when differentiated, appears elsewhere in the integrand.

For instance, in the given exercise, the substitution is made for the expression under the square root, turning it into a single variable, which makes integration more straightforward. After the substitution, the integral in terms of the new variable, usually noted as 'u', is solved, and the result is then substituted back in terms of the original variable.

Indefinite Integrals

Indefinite integrals represent the antiderivative of a function and are an essential part of integral calculus. Unlike definite integrals, indefinite integrals do not evaluate the function between limits. Instead, they provide a general form of the antiderivative, plus a constant term denoted as 'C', which represents all real constants.

The solution to our problem exemplifies this where the final result is expressed as an antiderivative of the integrated function added to a constant 'C'. It is essential for students to remember to always include the +C to account for any constant that was differentiated to a value of zero.

Integral Calculus Techniques

Integral calculus is rich with various techniques to tackle complex integrals, each with its own scenarios of applicability. Some common strategies beyond substitution include integration by parts, partial fractions, and trigonometric substitution. Mastery of these techniques allows for fluid transition between methods depending on the problem at hand.

The technique used in our example—substitution—was chosen upon recognizing that the differentiation of the lower function produced a simpler form that mirrored the upper function. This scenario is ideal for substitution, but other problems might require different approaches, highlighting the importance of a broad understanding of integration techniques.

Square Root Integration

Integration involving square roots can pose significant challenges, primarily because square root functions can result in non-polynomial integrands. To integrate functions with square roots, one may use substitution, as was done in the exercise, or other techniques like completing the square or trigonometric substitution, depending on the format of the square root expression.

In our exercise, by using substitution, the square root expression became a simpler power of the substitute variable 'u', which was then integrated using power rule. Students must familiarize themselves with these methods as square root integrals are common in both theoretical and applied mathematics.