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Chapter 7: Problem 27

Use a table of integrals to determine the following indefinite integrals. These integrals require preliminary work, such as completing the square or changing variables, before they can be found in a table. $$\int \frac{d x}{x\left(x^{10}+1\right)}$$

Short Answer

Expert verified
Question: Evaluate the indefinite integral: $$\int \frac{d x}{x\left(x^{10}+1\right)}.$$ Answer: The indefinite integral is $$\int \frac{d x}{x\left(x^{10}+1\right)} = \frac{1}{\sqrt{2}} \arctanh{\frac{x^5+\frac{1}{x^5}}{\sqrt{2}}} + C.$$

Step by step solution

01

Completing the square for the denominator

Let's first rewrite the polynomial \(x^{10} +1\) in terms of squares. To do this, we will complete the square. Recall that \((a+b)^2 = a^2 + 2ab + b^2\). In our case, let $$a = x^5$$ and $$b = \frac{1}{x^5}.$$ Therefore, we can rewrite the polynomial as: $$x^{10} + 1 = \left(x^5 + \frac{1}{x^5}\right)^2 - 2.$$
02

Substitution of variable

Now let's make a substitution to simplify the expression further: $$u = x^5 + \frac{1}{x^5}$$ which implies: $$\frac{du}{dx} = 5x^4 - \frac{5}{x^6}.$$ To find \(\frac{dx}{du}\), simply multiply both sides by \(\frac{dx}{du}=x^6\): $$x^6\frac{du}{dx} = 5x^{10} - 5.$$Divide by \(x^6\):$$\frac{du}{dx}=\frac{5x^{10} - 5}{x^6}.$$Now, divide by 5 and take the reciprocal of both sides to find \(\frac{dx}{du}\):$$\frac{dx}{du}=\frac{x^6}{x^{10}-1}.$$We are now ready to substitute \(u\) back into our integral and change variables.
03

Substituting back into the integral

We will now substitute \(u\) and \(\frac{dx}{du}\) back into our original integral and simplify: $$\int \frac{d x}{x\left(x^{10}+1\right)}=\int \frac{1}{x(u^2-2)}\left(\frac{x^6}{x^{10}-1}\right) du.$$We see that \(x^5\) cancels out, leaving us with:$$\int \frac{1}{u^2-2} du.$$Now we have an integral that is in a standard form that we can look it up in the table of integrals.
04

Using a table of integrals

The integral we have now is in the form of a rational function:$$\int \frac{1}{u^2-2} du.$$By referring to a table of integrals, we can find that this integral is equivalent to the following standard integral:$$\frac{1}{\sqrt{2}} \int \frac{1}{u^2-2} du = \frac{1}{\sqrt{2}} \arctanh{\frac{u}{\sqrt{2}}} + C. $$where C is the constant of integration.
05

Substituting back the original variable

Finally, let's substitute \(x^5 + \frac{1}{x^5}\) back in for \(u\) in our result:$$\frac{1}{\sqrt{2}} \arctanh{\frac{x^5+\frac{1}{x^5}}{\sqrt{2}}} + C.$$So the final result for the indefinite integral is:$$\int \frac{d x}{x\left(x^{10}+1\right)} = \frac{1}{\sqrt{2}} \arctanh{\frac{x^5+\frac{1}{x^5}}{\sqrt{2}}} + C.$$

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Most popular questions from this chapter

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