Question

Verify the following indefinite integrals by differentiation. These integrals are derived in later chapters. $$\int \frac{x}{\left(x^{2}-1\right)^{2}} d x=-\frac{1}{2\left(x^{2}-1\right)}+C$$

Short answer

Question: Verify the given indefinite integral by using differentiation: $$\int \frac{x}{\left(x^{2}-1\right)^{2}} d x=-\frac{1}{2\left(x^{2}-1\right)}+C$$ Solution: By differentiating the given expression with respect to x, we found out that: $$\frac{d}{dx}\left(-\frac{1}{2\left(x^{2}-1\right)}+C\right) = \frac{x}{\left(x^{2}-1\right)^{2}}$$ As this matches the given integrand, we can confirm that the given indefinite integral is correct.

Step-by-step solution

Differentiate the expression with respect to x

Using the power rule and chain rule, we differentiate the given expression. We differentiate each term independently since differentiation follows the linearity property. $$\frac{d}{dx}\left(-\frac{1}{2\left(x^{2}-1\right)}+C\right) = \frac{d}{dx}\left(-\frac{1}{2\left(x^{2}-1\right)}\right) + \frac{d}{dx}C$$

Differentiate the constant term, C

Recall that the derivative of any constant is zero. $$\frac{d}{dx}C = 0$$ So, $$\frac{d}{dx}\left(-\frac{1}{2\left(x^{2}-1\right)}+C\right) = \frac{d}{dx}\left(-\frac{1}{2\left(x^{2}-1\right)}\right)$$

Differentiate the non-constant term using the chain rule

Recall the chain rule: if y = u(v(x)), then: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$ Here, \(u(x) = 2(x^2 - 1)\) and \(v(x) = -\frac{1}{2}\). So, we have $$\frac{d}{dx}\left(-\frac{1}{2\left(x^{2}-1\right)}\right) = -\frac{1}{2} \cdot \frac{d}{dx}\left(\frac{1}{\left(x^{2}-1\right)}\right)$$ Let's differentiate the term inside brackets using the chain rule: $$\frac{d}{dx}\left(\frac{1}{\left(x^{2}-1\right)}\right) = -\frac{d}{dx}\left(\left(x^{2}-1\right)^{-1}\right) = -\left(x^{2}-1\right)^{-2} \cdot \frac{d}{dx}\left(x^{2}-1\right)$$ Using the power rule, we find: $$\frac{d}{dx}\left(x^{2}-1\right) = 2x$$ So, $$\frac{d}{dx}\left(\frac{1}{\left(x^{2}-1\right)}\right) = -2x\left(x^{2}-1\right)^{-2}$$ Now, plug this back into our original derivative expression: $$\frac{d}{dx}\left(-\frac{1}{2\left(x^{2}-1\right)}\right) = -\frac{1}{2} \cdot \left[-2x\left(x^{2}-1\right)^{-2}\right]$$

Simplify the derivative expression

$$\frac{d}{dx}\left(-\frac{1}{2\left(x^{2}-1\right)}\right) = x\left(x^{2}-1\right)^{-2} = \frac{x}{\left(x^{2}-1\right)^{2}}$$

Confirm that the derivative matches the integrand

We have found out that: $$\frac{d}{dx}\left(-\frac{1}{2\left(x^{2}-1\right)}+C\right) = \frac{x}{\left(x^{2}-1\right)^{2}}$$ As this matches the given integrand: $$\int \frac{x}{\left(x^{2}-1\right)^{2}} dx = -\frac{1}{2\left(x^{2}-1\right)}+C$$ Therefore, we can confirm that the given indefinite integral is correct.

Key concepts

Differentiation by First Principles

Understanding differentiation by first principles is akin to grasping the very foundations of calculus. This fundamental concept concerns the calculation of the derivative of a function from the ground up. In essence, it defines the derivative of a function at a point as the limit of the ratio of the change in the function's value to the change in the input value, as the latter approaches zero.

Using the definition of a limit, for a function f(x), the derivative f'(x) is given by: \[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]
The application of this principle can be quite laborious, as it requires the manipulation of this limit for different function formulas. However, it’s the origin from which all rules of differentiation are derived. So, when students verify indefinite integrals by differentiation, they're indirectly relying on the first principles even if they're using shortcuts like the power or the chain rule.

Power Rule

One of the shortcuts that streamlines differentiation is the power rule, which can be considered a direct descendant of the first principles. It applies to functions where the variable is raised to a power. If you have a function in the form of \(f(x) = x^n\), where n is any real number, then the power rule states that:

\[f'(x) = nx^{n-1}\]
For example, applying the power rule to \(x^2\), we get \(2x^1\) or simply \(2x\). It's incredibly efficient for polynomial functions and is often used in tandem with other rules for more complex functions. Our step-by-step solution showcases the power rule in action when differentiating \(x^2\) as part of a compound expression within the integrand.

Chain Rule

The chain rule is a vital technique in calculus, especially when dealing with the composition of functions. It allows you to differentiate composite functions where one function is nested inside another. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

To express this formally, if we have a composite function y = g(f(x)), then the chain rule tells us that:
\[\frac{dy}{dx} = \frac{dg}{df} \cdot \frac{df}{dx}\]
In the exercise provided, the chain rule is employed to differentiate the term \(-\frac{1}{2}(x^{2} - 1)^{-1}\), which is a composition of functions. The outer function is \(u^{-1}\), and the inner function is \(u = x^{2} - 1\). The chain rule simplifies the differentiation process and is instrumental in verifying the correctness of indefinite integrals, as we observed in our step-by-step solution.