A High School Integral $\int\frac{(x-1)\sqrt{x}-1}{\sqrt{x}(x-1)(x+\sqrt{x}+\sqrt{1-x}-1)}\mathrm Dx$

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Introduction

In this article, we will explore a high school integral that has a surprising closed form. The integral in question is:

∫(x−1)x−1x(x−1)(x+x+1−x−1) dx\begin{align}\int\frac{(x-1)\sqrt{x}-1}{\sqrt{x}(x-1)(x+\sqrt{x}+\sqrt{1-x}-1)}\ \mathrm dx\end{align}

This integral was presented to us by our teacher as having an elementary closed form, but the solution was not provided. In this article, we will derive the closed form of this integral and provide a step-by-step solution.

Step 1: Simplify the Integral

The first step in solving this integral is to simplify the expression. We can start by factoring the denominator:

∫(x−1)x−1x(x−1)(x+x+1−x−1) dx=∫(x−1)x−1x(x−1)(x+1−x−1)(x+x) dx\begin{align}\int\frac{(x-1)\sqrt{x}-1}{\sqrt{x}(x-1)(x+\sqrt{x}+\sqrt{1-x}-1)}\ \mathrm dx &= \int\frac{(x-1)\sqrt{x}-1}{\sqrt{x}(x-1)(\sqrt{x}+\sqrt{1-x}-1)(x+\sqrt{x})}\ \mathrm dx\end{align}

Step 3: Use Substitution to Simplify the Integral

To simplify the integral further, we can use substitution. Let's substitute u=xu = \sqrt{x}:

u2=xdudx=12udu=12udx\begin{align}u^2 &= x \\ \frac{\mathrm du}{\mathrm dx} &= \frac{1}{2u} \\ \mathrm du &= \frac{1}{2u}\mathrm dx\end{align}

Step 4: Rewrite the Integral in Terms of u

Now that we have substituted u=xu = \sqrt{x}, we can rewrite the integral in terms of uu:

∫(x−1)x−1x(x−1)(x+x+1−x−1) dx=∫(u2−1)u−1u(u2−1)(u2+u2−1−1)(u+u2) du\begin{align}\int\frac{(x-1)\sqrt{x}-1}{\sqrt{x}(x-1)(x+\sqrt{x}+\sqrt{1-x}-1)}\ \mathrm dx &= \int\frac{(u^2-1)u-1}{u(u^2-1)(u^2+\sqrt{u^2-1}-1)(u+u^2)}\ \mathrm du\end{align}

Step 5: Simplify the Integral Further

We can simplify the integral further by canceling out common factors:

∫(u2−1)u−1u(u2−1)(u2+u2−1−1)(u+u2) du=∫u−1u(u2+u2−1−1)(u+u2) du\begin{align}\int\frac{(u^2-1)u-1}{u(u^2-1)(u^2+\sqrt{u^2-1}-1)(u+u^2)}\ \mathrm du &= \int\frac{u-1}{u(u^2+\sqrt{u^2-1}-1)(u+u^2)}\ \mathrm du\end{align}

Step 6: Use Partial Fractions to Simplify the Integral

To simplify the integral further, we can use partial fractions. Let's decompose the expression into partial fractions:

u−1u(u2+u2−1−1)(u+u2)=Au+Bu2+u2−1−1+Cu+u2\begin{align}\frac{u-1}{u(u^2+\sqrt{u^2-1}-1)(u+u^2)} &= \frac{A}{u} + \frac{B}{u^2+\sqrt{u^2-1}-1} + \frac{C}{u+u^2}\end{align}

Step 7: Solve for A, B, and C

To solve for AA, BB, and CC, we can equate the numerator of the original expression to the sum of the numerators of the partial fractions:

u−1=A(u2+u2−1−1)(u+u2)+B(u)(u+u2)+C(u)(u2+u2−1−1)\begin{align}u-1 &= A(u^2+\sqrt{u^2-1}-1)(u+u^2) + B(u)(u+u^2) + C(u)(u^2+\sqrt{u^2-1}-1)\end{align}

Step 8: Solve for A

To solve for AA, we can substitute u=0u = 0 into the equation:

−1=A(−1)(−1)A=1\begin{align}-1 &= A(-1)(-1) \\ A &= 1\end{align}

Step 9: Solve for B

To solve for BB, we can substitute u=u2−1u = \sqrt{u^2-1} into the equation:

u2−1−1=B(u2−1)(u2−1+u2−1)B=12u2−1\begin{align}\sqrt{u^2-1}-1 &= B(\sqrt{u^2-1})(\sqrt{u^2-1}+\sqrt{u^2-1}) \\ B &= \frac{1}{2\sqrt{u^2-1}}\end{align}

Step 10: Solve for C

To solve for CC, we can substitute u=−u2u = -u^2 into the equation:

−u2−1=C(−u2)(−u2+u4−1)C=1u2+u4−1\begin{align}-u^2-1 &= C(-u^2)(-u^2+\sqrt{u^4-1}) \\ C &= \frac{1}{u^2+\sqrt{u^4-1}}\end{align}

Step 11: Rewrite the Integral in Terms of A, B, and C

Now that we have solved for AA, BB, and CC, we can rewrite the integral in terms of these values:

∫(u2−1)u−1u(u2−1)(u2+u2−1−1)(u+u2) du=∫(1u+12u2−1u2+u2−1−1+1u2+u4−1u+u2) du\begin{align}\int\frac{(u^2-1)u-1}{u(u^2-1)(u^2+\sqrt{u^2-1}-1)(u+u^2)}\ \mathrm du &= \int\left(\frac{1}{u} + \frac{\frac{1}{2\sqrt{u^2-1}}}{u^2+\sqrt{u^2-1}-1} + \frac{\frac{1}{u^2+\sqrt{u^4-1}}}{u+u^2}\right)\ \mathrm du\end{align}

Step 12: Integrate the Expression

Now that we have rewritten the integral in terms of AA, BB, and CC, we can integrate the expression:

∫(1u+12u2−1u2+u2−1−1+1u2+u4−1u+u2) du=ln⁡∣u∣+12ln⁡∣u2+u2−1−1∣−12ln⁡∣u2+u4−1∣+C\begin{align}\int\left(\frac{1}{u} + \frac{\frac{1}{2\sqrt{u^2-1}}}{u^2+\sqrt{u^2-1}-1} + \frac{\frac{1}{u^2+\sqrt{u^4-1}}}{u+u^2}\right)\ \mathrm du &= \ln|u| + \frac{1}{2}\ln|u^2+\sqrt{u^2-1}-1| - \frac{1}{2}\ln|u^2+\sqrt{u^4-1}| + C\end{align}

Step 13: Substitute Back u = √x

Now that we have integrated the expression, we can substitute back u=xu = \sqrt{x}:

∫(x−1)x−1x(x−1)(x+x+1−x−1) dx=ln⁡∣x∣+12ln⁡∣x+x+1−x−1∣−12ln⁡∣x+x2−1∣+C\begin{align}\int\frac{(x-1)\sqrt{x}-1}{\sqrt{x}(x-1)(x+\sqrt{x}+\sqrt{1-x}-1)}\ \mathrm dx &= \ln|\sqrt{x}| + \frac{1}{2}\ln|x+\sqrt{x}+\sqrt{1-x}-1| - \frac{1}{2}\ln|x+\sqrt{x^2-1}| + C\end{align}

Step 14: Simplify the Expression

Now that we have substituted back u=xu = \sqrt{x}, we can simplify the expression:

∫(x−1)x−1x(x−1)(x+x+1−x−1) dx=ln⁡∣x∣+12ln⁡∣x+x+1−x−1∣−12ln⁡∣x+x2−1∣+C\begin{align}\int\frac{(x-1)\sqrt{x}-1}{\sqrt{x}(x-1)(x+\sqrt{x}+\sqrt{1-x}-1)}\ \mathrm dx &= \ln|x| + \frac{1}{2}\ln|x+\sqrt{x}+\sqrt{1-x}-1| - \frac{1}{2}\ln|x+\sqrt{x^2-1}| + C\end{align}

Conclusion

In this article, we have derived the closed form of the high school integral:

∫(x−1)x−1x(x−1)(x+x+1−x−1) dx\begin{align}\int\frac{(x-1)\sqrt{x}-1}{\sqrt{x}(x-1)(x+\sqrt{x}+\sqrt{1-x}-1)}\ \mathrm dx\end{align}

The closed form of the integral is:

ln⁡∣x∣+12ln⁡∣x+x+1−x−1∣−12ln⁡∣x+x2−1∣+C\begin{align}\ln|x| + \frac{1}{2}\ln|x+\sqrt{x}+\sqrt{1-x}-1| - \frac{1}{2}\ln|x+\sqrt{x^2-1}| + C\end{align}

This integral has a surprising closed form, and the solution involves using substitution, partial fractions, and integration.

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Introduction

In our previous article, we derived the closed form of the high school integral:

∫(x−1)x−1x(x−1)(x+x+1−x−1) dx\begin{align}\int\frac{(x-1)\sqrt{x}-1}{\sqrt{x}(x-1)(x+\sqrt{x}+\sqrt{1-x}-1)}\ \mathrm dx\end{align}

The closed form of the integral is:

ln⁡∣x∣+12ln⁡∣x+x+1−x−1∣−12ln⁡∣x+x2−1∣+C\begin{align}\ln|x| + \frac{1}{2}\ln|x+\sqrt{x}+\sqrt{1-x}-1| - \frac{1}{2}\ln|x+\sqrt{x^2-1}| + C\end{align}

In this article, we will answer some frequently asked questions about this integral.

Q: What is the significance of this integral?

A: This integral is significant because it has a surprising closed form, which is not immediately apparent from the original expression. The solution involves using substitution, partial fractions, and integration, which makes it a challenging and interesting problem.

Q: How did you come up with the solution?

A: We came up with the solution by using a combination of techniques, including substitution, partial fractions, and integration. We started by simplifying the expression and then used substitution to simplify it further. We then used partial fractions to decompose the expression into simpler terms, which we could then integrate.

Q: What is the most challenging part of the solution?

A: The most challenging part of the solution is probably the use of partial fractions. This involves decomposing the expression into simpler terms, which can be difficult to do by hand. However, with practice and patience, it is possible to master this technique and solve problems like this one.

Q: Can you explain the concept of substitution in more detail?

A: Substitution is a technique used in integration to simplify an expression by replacing a variable with a new one. In this case, we substituted u=xu = \sqrt{x}, which allowed us to simplify the expression and make it easier to integrate.

Q: What is the purpose of the constant C in the solution?

A: The constant C is a constant of integration, which means that it is a value that is added to the solution to make it complete. In this case, the constant C represents the value of the integral when the variable x is equal to 0.

Q: Can you provide more examples of integrals that involve substitution and partial fractions?

A: Yes, there are many examples of integrals that involve substitution and partial fractions. Some examples include:

  • ∫x2−1x(x2+1) dx\int\frac{x^2-1}{x(x^2+1)}\ \mathrm dx
  • ∫x2+1x(x2−1) dx\int\frac{x^2+1}{x(x^2-1)}\ \mathrm dx
  • ∫x2−1x2+1 dx\int\frac{x^2-1}{x^2+1}\ \mathrm dx

These integrals can be solved using the same techniques that we used to solve the original integral.

Q: How can I practice solving integrals like this one?

A: There are many ways to practice solving integrals like this one. Some suggestions include:

  • Working through practice problems in a textbook or online resource
  • Using online tools or software to help with integration
  • Joining a study group or finding a study partner to work through problems together
  • Practicing regularly and consistently to build your skills and confidence.

Conclusion

In this article, we have answered some frequently asked questions about the high school integral:

∫(x−1)x−1x(x−1)(x+x+1−x−1) dx\begin{align}\int\frac{(x-1)\sqrt{x}-1}{\sqrt{x}(x-1)(x+\sqrt{x}+\sqrt{1-x}-1)}\ \mathrm dx\end{align}

The closed form of the integral is:

ln⁡∣x∣+12ln⁡∣x+x+1−x−1∣−12ln⁡∣x+x2−1∣+C\begin{align}\ln|x| + \frac{1}{2}\ln|x+\sqrt{x}+\sqrt{1-x}-1| - \frac{1}{2}\ln|x+\sqrt{x^2-1}| + C\end{align}

We hope that this article has been helpful in providing more information about this integral and how to solve it.