Evaluate ∫ D X ( X + 1 ) 2 ( X − 1 ) 4 3 \int\frac{dx}{\sqrt[3]{(x+1)^2(x-1)^4}} ∫ 3 ( X + 1 ) 2 ( X − 1 ) 4 ​ D X ​

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In the realm of calculus, the evaluation of integrals stands as a fundamental skill, often requiring a blend of algebraic manipulation, strategic substitutions, and a keen understanding of integration techniques. The integral presented, $\int\frac{dx}{\sqrt[3]{(x+1)2(x-1)4}}$, exemplifies a challenging yet rewarding problem that invites exploration through various methods. This comprehensive guide delves into the intricacies of solving this integral, offering a step-by-step approach, insightful discussions on the techniques employed, and a focus on optimizing the solution for clarity and precision. Let's embark on this journey together, unraveling the complexities and arriving at a satisfying resolution.

Initial Assessment and Strategic Planning

Before diving into the solution, it's crucial to assess the integral and devise a strategic plan. The presence of a radical expression in the denominator, specifically a cube root containing polynomial factors, suggests that a u-substitution or a trigonometric substitution might be beneficial. However, the specific form of the expression, with different powers of $(x+1)$ and $(x-1)$, hints that a more tailored substitution could lead to a more elegant solution. The provided hint, suggesting a substitution of the form $\sqrt[n]{\frac{ax+b}{cx+d}}=t$, is a valuable clue. This substitution aims to rationalize the expression within the cube root, potentially simplifying the integral significantly.

The key to successful integration often lies in recognizing patterns and choosing the appropriate technique. In this case, the fractional powers and the structure of the integrand strongly suggest that a rationalizing substitution will be the most effective approach. We'll explore how this substitution works in detail, transforming the integral into a more manageable form.

Employing the Rationalizing Substitution

The suggested substitution, $\sqrt[n]{\frac{ax+b}{cx+d}}=t$, serves as a powerful tool for handling integrals with radical expressions. In our specific case, we have a cube root, which means we're dealing with $n=3$. The expression inside the cube root can be rearranged to fit the form $\frac{ax+b}{cx+d}$. Let's rewrite the integrand to highlight this structure:

$$\int\frac{dx}{\sqrt[3]{(x+1)^2(x-1)^4}} = \int\frac{dx}{\sqrt[3]{(x+1)^2(x-1)^3(x-1)}} = \int\frac{dx}{(x-1)\sqrt[3]{\frac{(x+1)^2}{(x-1)^2}}}$$

Now, we can see the fraction inside the cube root more clearly. To simplify this, let's make the substitution:

$$t = \sqrt[3]{\frac{x+1}{x-1}}$$

This substitution is a clever way to rationalize the cube root, aiming to eliminate the radical and simplify the integral. To proceed, we need to express $x$ and $dx$ in terms of $t$. First, cube both sides:

$$t^3 = \frac{x+1}{x-1}$$

Now, solve for $x$:

$$t^3(x-1) = x+1$$

$$t^3x - t^3 = x+1$$

$$t^3x - x = t^3 + 1$$

$$x(t^3 - 1) = t^3 + 1$$

$$x = \frac{t^3 + 1}{t^3 - 1}$$

Next, we need to find $dx$ in terms of $dt$. Differentiate both sides of the equation $x = \frac{t^3 + 1}{t^3 - 1}$ with respect to $t$:

$$\frac{dx}{dt} = \frac{(3t^2)(t^3 - 1) - (t^3 + 1)(3t^2)}{(t^3 - 1)^2}$$

$$\frac{dx}{dt} = \frac{3t^5 - 3t^2 - 3t^5 - 3t^2}{(t^3 - 1)^2}$$

$$\frac{dx}{dt} = \frac{-6t^2}{(t^3 - 1)^2}$$

Therefore,

$$dx = \frac{-6t^2}{(t^3 - 1)^2} dt$$

Now we have both $x$ and $dx$ in terms of $t$, ready to be substituted back into the original integral.

Substituting and Simplifying the Integral

With the substitution $t = \sqrt[3]{\frac{x+1}{x-1}}$, $x = \frac{t^3 + 1}{t^3 - 1}$, and $dx = \frac{-6t^2}{(t^3 - 1)^2} dt$, we can now substitute these expressions into the integral. Recall the rewritten form of the integral:

$$\int\frac{dx}{(x-1)\sqrt[3]{\frac{(x+1)^2}{(x-1)^2}}}$$

Let's substitute $t$ into the cube root part:

$$\sqrt[3]{\frac{(x+1)^2}{(x-1)^2}} = \left(\sqrt[3]{\frac{x+1}{x-1}}\right)^2 = t^2$$

Now we need to express $(x-1)$ in terms of $t$. Using the expression for $x$:

$$x - 1 = \frac{t^3 + 1}{t^3 - 1} - 1 = \frac{t^3 + 1 - (t^3 - 1)}{t^3 - 1} = \frac{2}{t^3 - 1}$$

Now we can substitute everything back into the integral:

$$\int\frac{dx}{(x-1)\sqrt[3]{\frac{(x+1)^2}{(x-1)^2}}} = \int \frac{\frac{-6t^2}{(t^3 - 1)^2} dt}{\left(\frac{2}{t^3 - 1}\right)t^2}$$

Simplify the expression:

$$\int \frac{-6t^2}{(t^3 - 1)^2} \cdot \frac{t^3 - 1}{2} \cdot \frac{1}{t^2} dt = \int \frac{-3}{t^3 - 1} dt$$

This substitution has transformed the integral into a simpler, rational form. The next step involves integrating this expression, which often requires partial fraction decomposition.

Applying Partial Fraction Decomposition

Now we need to evaluate the integral:

$$\int \frac{-3}{t^3 - 1} dt$$

The denominator, $t^3 - 1$, can be factored as $(t-1)(t^2 + t + 1)$. This suggests using partial fraction decomposition. We want to express the integrand as:

$$\frac{-3}{t^3 - 1} = \frac{A}{t-1} + \frac{Bt + C}{t^2 + t + 1}$$

Multiply both sides by $t^3 - 1$ to clear the denominators:

$$-3 = A(t^2 + t + 1) + (Bt + C)(t - 1)$$

Expand the expression:

$$-3 = At^2 + At + A + Bt^2 - Bt + Ct - C$$

Group the terms by powers of $t$:

$$-3 = (A + B)t^2 + (A - B + C)t + (A - C)$$

Now we can equate the coefficients of the corresponding powers of $t$:

• $A + B = 0$ (coefficient of $t^2$)
• $A - B + C = 0$ (coefficient of $t$)
• $A - C = -3$ (constant term)

From the first equation, we have $B = -A$. Substituting this into the second equation:

$$A - (-A) + C = 0$$

$$2A + C = 0$$

Now we have a system of two equations:

• $2A + C = 0$
• $A - C = -3$

Add the two equations:

$$3A = -3$$

$$A = -1$$

Now we can find $B$ and $C$:

$$B = -A = 1$$

$$C = A + 3 = -1 + 3 = 2$$

So we have $A = -1$, $B = 1$, and $C = 2$. Thus, the partial fraction decomposition is:

$$\frac{-3}{t^3 - 1} = \frac{-1}{t-1} + \frac{t + 2}{t^2 + t + 1}$$

Now we can rewrite the integral as:

$$\int \frac{-3}{t^3 - 1} dt = \int \left(\frac{-1}{t-1} + \frac{t + 2}{t^2 + t + 1}\right) dt$$

This integral can now be split into two simpler integrals.

Integrating the Decomposed Fractions

We now have the integral decomposed into two parts:

$$\int \left(\frac{-1}{t-1} + \frac{t + 2}{t^2 + t + 1}\right) dt = -\int \frac{1}{t-1} dt + \int \frac{t + 2}{t^2 + t + 1} dt$$

The first integral is straightforward:

$$- \int \frac{1}{t-1} dt = -\ln|t-1| + C_1$$

The second integral requires a bit more work. We'll complete the square in the denominator:

$$t^2 + t + 1 = \left(t + \frac{1}{2}\right)^2 + \frac{3}{4}$$

Now, rewrite the integral:

$$\int \frac{t + 2}{t^2 + t + 1} dt = \int \frac{t + 2}{\left(t + \frac{1}{2}\right)^2 + \frac{3}{4}} dt$$

Let $u = t + \frac{1}{2}$, so $t = u - \frac{1}{2}$ and $du = dt$. The integral becomes:

$$\int \frac{u - \frac{1}{2} + 2}{u^2 + \frac{3}{4}} du = \int \frac{u + \frac{3}{2}}{u^2 + \frac{3}{4}} du$$

Split this integral into two parts:

$$\int \frac{u}{u^2 + \frac{3}{4}} du + \int \frac{\frac{3}{2}}{u^2 + \frac{3}{4}} du$$

The first integral can be solved with a simple substitution: let $v = u^2 + \frac{3}{4}$, so $dv = 2u du$:

$$\int \frac{u}{u^2 + \frac{3}{4}} du = \frac{1}{2} \int \frac{dv}{v} = \frac{1}{2} \ln|v| + C_2 = \frac{1}{2} \ln\left(u^2 + \frac{3}{4}\right) + C_2$$

The second integral is an arctangent integral:

$$\frac{3}{2} \int \frac{1}{u^2 + \frac{3}{4}} du = \frac{3}{2} \cdot \frac{2}{\sqrt{3}} \arctan\left(\frac{u}{\frac{\sqrt{3}}{2}}\right) + C_3 = \sqrt{3} \arctan\left(\frac{2u}{\sqrt{3}}\right) + C_3$$

Combine these results:

$$\int \frac{t + 2}{t^2 + t + 1} dt = \frac{1}{2} \ln\left(\left(t + \frac{1}{2}\right)^2 + \frac{3}{4}\right) + \sqrt{3} \arctan\left(\frac{2(t + \frac{1}{2})}{\sqrt{3}}\right) + C_2 + C_3$$

$$\int \frac{t + 2}{t^2 + t + 1} dt = \frac{1}{2} \ln(t^2 + t + 1) + \sqrt{3} \arctan\left(\frac{2t + 1}{\sqrt{3}}\right) + C_4$$

Now we can combine the results from both decomposed integrals:

$$\int \frac{-3}{t^3 - 1} dt = -\ln|t-1| + \frac{1}{2} \ln(t^2 + t + 1) + \sqrt{3} \arctan\left(\frac{2t + 1}{\sqrt{3}}\right) + C$$

Back-Substitution and Final Solution

Finally, we need to substitute back $t = \sqrt[3]{\frac{x+1}{x-1}}$ into the result:

$$\int\frac{dx}{\sqrt[3]{(x+1)^2(x-1)^4}} = -\ln\left|\sqrt[3]{\frac{x+1}{x-1}} - 1\right| + \frac{1}{2} \ln\left(\left(\sqrt[3]{\frac{x+1}{x-1}}\right)^2 + \sqrt[3]{\frac{x+1}{x-1}} + 1\right) + \sqrt{3} \arctan\left(\frac{2\sqrt[3]{\frac{x+1}{x-1}} + 1}{\sqrt{3}}\right) + C$$

This is the final solution to the integral. It may look complex, but it's a precise representation of the antiderivative. The process involved a clever rationalizing substitution, partial fraction decomposition, and careful integration of the resulting terms. This example showcases the power and versatility of various integration techniques in calculus.

Conclusion: Mastering Integration Techniques

Evaluating the integral $\int\frac{dx}{\sqrt[3]{(x+1)^2(x-1)^4}}$ has been a journey through several key calculus concepts. We began by assessing the integral and devising a strategic plan, recognizing the need for a rationalizing substitution. The substitution $t = \sqrt[3]{\frac{x+1}{x-1}}$ proved to be a powerful tool, transforming the integral into a more manageable form. From there, we employed partial fraction decomposition to break down the rational function into simpler terms, and finally, we integrated each term, carefully handling the algebraic manipulations and trigonometric substitutions required.

The final solution, while complex in appearance, represents the accurate antiderivative of the original integrand. This exercise underscores the importance of mastering various integration techniques and developing a strategic approach to problem-solving in calculus. The ability to recognize patterns, choose appropriate substitutions, and apply algebraic manipulations with precision is crucial for success in integral calculus. This comprehensive guide has not only provided a solution to this specific integral but also illuminated the broader principles of integral evaluation, empowering readers to tackle similar challenges with confidence and competence. As you continue your exploration of calculus, remember that practice and a deep understanding of fundamental concepts are the keys to unlocking the beauty and power of integration.