A High School Integral ∫ X ( 2 X + 3 ) X + 1 3 X 6 + X 2 + 2 X + 1 D X \int\frac{x (2x + 3) \sqrt[3]{x + 1}}{x^6 + X^2 + 2x + 1}\,\mathrm Dx ∫ X 6 + X 2 + 2 X + 1 X ( 2 X + 3 ) 3 X + 1 ​ ​ D X

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Introduction

In this article, we will delve into the evaluation of a high school integral that has a surprising closed form. The integral in question is $\int\frac{x (2x + 3) \sqrt[3]{x + 1}}{x^6 + x^2 + 2x + 1},\mathrm dx.$ This integral has been assigned to students in high school calculus classes, and it is often considered a challenging problem. However, as we will see, the integral has a surprisingly simple closed form.

Background

Before we begin evaluating the integral, let's take a closer look at the denominator. The denominator is a polynomial of degree 6, which can be factored as follows:

x6+x2+2x+1=(x2+1)3−(x2+1)+1.x^6 + x^2 + 2x + 1 = (x^2 + 1)^3 - (x^2 + 1) + 1.

This factorization is not immediately obvious, but it is a useful one. We will see why later.

Evaluating the Integral

To evaluate the integral, we will use a combination of algebraic manipulations and integration by parts. We start by rewriting the integral as follows:

∫x(2x+3)x+13x6+x2+2x+1 dx=∫x(2x+3)x+13(x2+1)3−(x2+1)+1 dx.\int\frac{x (2x + 3) \sqrt[3]{x + 1}}{x^6 + x^2 + 2x + 1}\,\mathrm dx = \int\frac{x (2x + 3) \sqrt[3]{x + 1}}{(x^2 + 1)^3 - (x^2 + 1) + 1}\,\mathrm dx.

Next, we make the substitution u=x2+1u = x^2 + 1. This gives us du=2x dxdu = 2x\,dx, which we can use to rewrite the integral as follows:

∫x(2x+3)x+13(x2+1)3−(x2+1)+1 dx=∫(u−1)(2(u−1)+3)u−13u3−u+1 du.\int\frac{x (2x + 3) \sqrt[3]{x + 1}}{(x^2 + 1)^3 - (x^2 + 1) + 1}\,\mathrm dx = \int\frac{(u - 1) (2(u - 1) + 3) \sqrt[3]{u - 1}}{u^3 - u + 1}\,\mathrm du.

Integration by Parts

We will now use integration by parts to evaluate the integral. We choose u=(u−1)(2(u−1)+3)u−13u = (u - 1) (2(u - 1) + 3) \sqrt[3]{u - 1} and dv=1u3−u+1 dudv = \frac{1}{u^3 - u + 1}\,du. This gives us du=(2u−1)(2(u−1)+3)u−13+(u−1)(2(u−1)+3)13(u−1)2/3 dudu = (2u - 1) (2(u - 1) + 3) \sqrt[3]{u - 1} + (u - 1) (2(u - 1) + 3) \frac{1}{3(u - 1)^{2/3}}\,du and v=∫1u3−u+1 duv = \int\frac{1}{u^3 - u + 1}\,du.

Evaluating the Integral of 1/(u^3-u+1)

The integral of 1u3−u+1\frac{1}{u^3 - u + 1} can be evaluated using partial fractions. We can write 1u3−u+1=Au−1+Bu2+u+1\frac{1}{u^3 - u + 1} = \frac{A}{u - 1} + \frac{B}{u^2 + u + 1} for some constants AA and BB. We can then use the method of equating coefficients to find the values of AA and BB.

Finding the Values of A and B

We can equate the coefficients of the u2u^2 term on both sides of the equation to get A=−13A = -\frac{1}{3}. We can then equate the constant terms on both sides of the equation to get B=13B = \frac{1}{3}.

Evaluating the Integral

We can now use the values of AA and BB to evaluate the integral of 1u3−u+1\frac{1}{u^3 - u + 1}. We have

∫1u3−u+1 du=−13∫1u−1 du+13∫1u2+u+1 du.\int\frac{1}{u^3 - u + 1}\,du = -\frac{1}{3}\int\frac{1}{u - 1}\,du + \frac{1}{3}\int\frac{1}{u^2 + u + 1}\,du.

Evaluating the Integrals

We can evaluate the integrals on the right-hand side of the equation using the method of substitution. We have

∫1u−1 du=ln⁡∣u−1∣+C1\int\frac{1}{u - 1}\,du = \ln|u - 1| + C_1

and

∫1u2+u+1 du=ln⁡∣u2+u+1∣+C2.\int\frac{1}{u^2 + u + 1}\,du = \ln|u^2 + u + 1| + C_2.

Combining the Results

We can now combine the results to get

∫1u3−u+1 du=−13ln⁡∣u−1∣+13ln⁡∣u2+u+1∣+C.\int\frac{1}{u^3 - u + 1}\,du = -\frac{1}{3}\ln|u - 1| + \frac{1}{3}\ln|u^2 + u + 1| + C.

Back to the Original Integral

We can now substitute back to get

∫x(2x+3)x+13x6+x2+2x+1 dx=13ln⁡∣(x2+1)2+(x2+1)+1x2+1−1∣+C.\int\frac{x (2x + 3) \sqrt[3]{x + 1}}{x^6 + x^2 + 2x + 1}\,\mathrm dx = \frac{1}{3}\ln\left|\frac{(x^2 + 1)^2 + (x^2 + 1) + 1}{x^2 + 1 - 1}\right| + C.

Simplifying the Result

We can simplify the result to get

∫x(2x+3)x+13x6+x2+2x+1 dx=13ln⁡∣(x2+1)2+(x2+1)+1∣+C.\int\frac{x (2x + 3) \sqrt[3]{x + 1}}{x^6 + x^2 + 2x + 1}\,\mathrm dx = \frac{1}{3}\ln\left|(x^2 + 1)^2 + (x^2 + 1) + 1\right| + C.

Conclusion

In this article, we have evaluated the high school integral $\int\frac{x (2x + 3) \sqrt[3]{x + 1}}{x^6 + x^2 + 2x + 1},\mathrm dx.$ We have used a combination of algebraic manipulations and integration by parts to get a surprisingly simple closed form. The result is $\frac{1}{3}\ln\left|(x^2 + 1)^2 + (x^2 + 1) + 1\right| + C.$ This result is a great example of how careful algebraic manipulations and integration techniques can be used to evaluate seemingly difficult integrals.

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Introduction

In our previous article, we evaluated the high school integral $\int\frac{x (2x + 3) \sqrt[3]{x + 1}}{x^6 + x^2 + 2x + 1},\mathrm dx.$ We used a combination of algebraic manipulations and integration by parts to get a surprisingly simple closed form. In this article, we will answer some common questions that students may have when evaluating this integral.

Q: What is the significance of the denominator in the integral?

A: The denominator in the integral is a polynomial of degree 6, which can be factored as follows:

x6+x2+2x+1=(x2+1)3−(x2+1)+1.x^6 + x^2 + 2x + 1 = (x^2 + 1)^3 - (x^2 + 1) + 1.

This factorization is not immediately obvious, but it is a useful one. We used this factorization to make the substitution u=x2+1u = x^2 + 1, which simplified the integral.

Q: Why did we use integration by parts to evaluate the integral?

A: We used integration by parts to evaluate the integral because the integral of 1u3−u+1\frac{1}{u^3 - u + 1} can be evaluated using partial fractions. We can write 1u3−u+1=Au−1+Bu2+u+1\frac{1}{u^3 - u + 1} = \frac{A}{u - 1} + \frac{B}{u^2 + u + 1} for some constants AA and BB. We can then use the method of equating coefficients to find the values of AA and BB.

Q: How did we find the values of A and B?

A: We found the values of AA and BB by equating the coefficients of the u2u^2 term on both sides of the equation to get A=−13A = -\frac{1}{3}. We then equated the constant terms on both sides of the equation to get B=13B = \frac{1}{3}.

Q: What is the final answer to the integral?

A: The final answer to the integral is $\frac{1}{3}\ln\left|(x^2 + 1)^2 + (x^2 + 1) + 1\right| + C.$

Q: Why is the answer so complicated?

A: The answer is complicated because we used a combination of algebraic manipulations and integration by parts to evaluate the integral. However, the answer is still a surprisingly simple closed form.

Q: Can you give an example of how to use this integral in a real-world problem?

A: Yes, here is an example of how to use this integral in a real-world problem. Suppose we want to find the volume of a solid that is formed by rotating a region about the x-axis. The region is bounded by the curves y=x+13y = \sqrt[3]{x + 1} and y=x(2x+3)x6+x2+2x+1y = \frac{x (2x + 3)}{x^6 + x^2 + 2x + 1}. We can use the integral we evaluated earlier to find the volume of the solid.

Q: How do we use the integral to find the volume of the solid?

A: We use the integral to find the volume of the solid by integrating the area of the region with respect to x. The area of the region is given by the integral we evaluated earlier, and we can use this integral to find the volume of the solid.

Q: What is the final answer to the volume of the solid?

A: The final answer to the volume of the solid is $\frac{1}{3}\ln\left|(x^2 + 1)^2 + (x^2 + 1) + 1\right| + C.$

Conclusion

In this article, we answered some common questions that students may have when evaluating the high school integral $\int\frac{x (2x + 3) \sqrt[3]{x + 1}}{x^6 + x^2 + 2x + 1},\mathrm dx.$ We used a combination of algebraic manipulations and integration by parts to get a surprisingly simple closed form. We also gave an example of how to use this integral in a real-world problem.