Question

Evaluate the following integrals : $$ \int \frac{\sqrt[3]{1+x^{3}}}{x^{2}} d x $$

Short answer

#Answer: $$ \int \frac{\sqrt[3]{1+x^{3}}}{x^{2}} dx = \frac{3}{7}(1 + x^3)^{\frac{7}{3}} - 3(1 + x^3)^{\frac{5}{3}} + \frac{3}{4}(1 + x^3)^{\frac{4}{3}} + C $$

Step-by-step solution

Choose the proper substitution

Choose \(u\) as: $$ u = \sqrt[3]{1 + x^3} $$ Now differentiate \(u\) with respect to \(x\): $$ \frac{du}{dx} = \frac{d}{dx}(\sqrt[3]{1 + x^3}) $$ Using the chain rule, we have: $$ \frac{du}{dx} = \frac{1}{3}(1 + x^3)^{-\frac{2}{3}}(3x^2) $$ Based on this, we can find \(dx\) in terms of \(du\): $$ dx = \frac{du}{(x^2(1 + x^3)^{-\frac{2}{3}})} $$

Rewrite the integral in terms of \(u\)

Let's rewrite the integral by replacing \(x\) with \(u\). So we have: $$ \int \frac{\sqrt[3]{1+x^{3}}}{x^{2}} dx = \int \frac{u}{x^2} \cdot \frac{du}{(x^2(1 + x^3)^{-\frac{2}{3}})} $$ Notice that \(x^2\) cancels out, and we have: $$ \int \frac{\sqrt[3]{1+x^{3}}}{x^{2}} dx = \int u(1 + x^3)^{\frac{2}{3}} du $$ Don't forget that we need to rewrite \((1+x^3)\) in terms of \(u\): $$ (1+x^3) = u^3 \implies x^3 = u^3 - 1 $$ With this substitution, our integral becomes: $$ \int \frac{\sqrt[3]{1+x^{3}}}{x^{2}} dx = \int u(u^3 - 1)^{\frac{2}{3}} du $$

Expand and evaluate the integral

Now, let's expand the integral and then evaluate it: $$ \int u(u^3 - 1)^{\frac{2}{3}} du = \int \left(u^4 - 2u^3 + u^2\right)^{\frac{2}{3}} du $$ This integral can be evaluated using power rule: $$ \int (u^4 - 2u^3 + u^2)^{\frac{2}{3}} du = \frac{3}{7}u^7 - {3u^5} + \frac{3}{4}u^4 + C $$

Replace \(u\) with \(x\)

Finally, revert back to the original variable \(x\): $$ \int \frac{\sqrt[3]{1+x^{3}}}{x^{2}} dx = \frac{3}{7}(\sqrt[3]{1 + x^3})^7 - 3(\sqrt[3]{1 + x^3})^5 + \frac{3}{4}(\sqrt[3]{1 + x^3})^4 + C $$ So, the evaluated integral is: $$ \int \frac{\sqrt[3]{1+x^{3}}}{x^{2}} dx = \frac{3}{7}(1 + x^3)^{\frac{7}{3}} - 3(1 + x^3)^{\frac{5}{3}} + \frac{3}{4}(1 + x^3)^{\frac{4}{3}} + C $$

Key concepts

U-Substitution

When integrating complex functions, u-substitution is a technique that can simplify the process significantly. It is the reverse process of the chain rule in differentiation and often referred to as a change of variables. When a function’s integrand is a composite function, especially where the derivative of the inner function is present elsewhere in the integrand, u-substitution can be a powerful tool.

To employ this method, you first identify a section of the integrand that you will assign to 'u'. Ideally, this is a part whose differential 'du' is also noticeable in the integrand. In our exercise, we chose u as \( u = \sqrt[3]{1 + x^3} \), because the differential of this function, which involves x, appears in the integrand.

After differentiating u with respect to x, you then rewrite the integral in terms of u, which sometimes simplifies the integrand or turns it into a form that is easier to integrate.

Chain Rule

The chain rule is a fundamental rule in calculus for differentiating compositions of functions. It expresses the derivative of a composite function in terms of the derivatives of its inner and outer functions. The general form for functions f(g(x)) is given by \( \frac{df}{dx} = \frac{df}{dg}\cdot\frac{dg}{dx} \).

In our integral calculus problem, we used the chain rule to differentiate \(u = \sqrt[3]{1 + x^3}\), resulting in \( \frac{du}{dx} = \frac{1}{3}(1 + x^3)^{-\frac{2}{3}}(3x^2) \), which is essential for finding \(dx\) in terms of \(du\). Understanding the chain rule is vital for correctly applying u-substitution when integrating functions.

Power Rule

The power rule is a quick and easy way to differentiate and integrate monomials. For differentiation, if you have a function \(x^n\), its derivative is \(nx^{n-1}\). The rule also applies in reverse for integration, with the antiderivative of \(x^n\) being \(\frac{1}{n+1}x^{n+1}\), plus an integration constant \(C\).

When we expanded and integrated our function in terms of u, the power rule allowed us to find the antiderivative of terms like \(u^4\) and \(u^2\), which ultimately led us to the answer \(\frac{3}{7}u^7 - 3u^5 + \frac{3}{4}u^4 + C\). It is important to note that the power rule for integration does not apply to functions of the form \(x^{-1}\), which integrate to \(\ln|x| + C\).

Definite Integral

A definite integral is an integral with upper and lower limits, which gives the net area under the curve of a function on a specific interval. It is represented as \(\int_{a}^{b} f(x)dx\), where 'a' and 'b' are the lower and upper limits, respectively. Unlike indefinite integrals, definite integrals result in a number and do not include an arbitrary constant \(C\).

While our example focuses on an indefinite integral, understanding definite integrals is helpful when dealing with real-world applications, such as calculating distances, areas, and volumes. Evaluating a definite integral involves finding the indefinite integral (antiderivative) first and then applying the limits to the result. This process is often facilitated by the Fundamental Theorem of Calculus, which links differentiation with the definite integral.