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Chapter 7: Problem 44

Evaluate the following integrals. $$\int \frac{d x}{x^{3} \sqrt{x^{2}-1}}, x>1$$

Short Answer

Expert verified
Question: Evaluate the integral $$\int \frac{dx}{x^{3} \sqrt{x^{2}-1}}$$ Answer: The evaluated integral is $$\frac{1}{x^4}(\ln{(\sqrt{x^2-1})} - \arctan{(\sqrt{x^2-1})}) + C$$, where C is the constant of integration.

Step by step solution

01

Perform Substitution

Let's perform the substitution $$u = x^2$$, which implies that $$\frac{du}{dx} = 2x$$, or $$dx = \frac{du}{2x}$$. Now, we can rewrite the integral in terms of u: $$\int \frac{d x}{x^{3} \sqrt{x^{2}-1}} = \int \frac{1}{x^3 \sqrt{u-1}} \frac{du}{2x}$$
02

Simplify the Integral

Now, we can simplify the integral expression: $$\int \frac{du}{2x^4 \sqrt{u-1}}$$
03

Use Partial Fractions Decomposition

Since we are looking at an integral of the form \(\int \frac{du}{u^n(a-u)^n}\) (u^n and (a-u)^n have the same exponent), we can use partial fractions decomposition with the substitution $$v = \sqrt{u-1}$$ , so, $$u = v^2+1$$ and $$du = 2v dv$$: Now our integral becomes: $$\frac{1}{2}\int \frac{2v dv}{x^4(v^2 + 1) v}$$ $$\frac{1}{x^4}\int \frac{dv}{v^3+ v}$$
04

Evaluate the Integral

The integral can be now written in terms of v and it looks much easier: $$\frac{1}{x^4}\int \frac{dv}{v(v^2+1)}$$ Now, perform a partial fraction decomposition: $$\frac{A}{v} + \frac{Bv+C}{v^2+1}$$ Solve this system of equations, and you'll find A = 1, B = 0, and C = -1. So, the integral becomes: $$\frac{1}{x^4}\left(\int \frac{dv}{v} - \int \frac{dv}{v^2+1}\right)$$
05

Evaluate the Two Integrals

By splitting the integral and evaluating each, we get: $$\frac{1}{x^4}\left(\int \frac{dv}{v} - \int \frac{dv}{v^2+1}\right) = \frac{1}{x^4}(\ln{v} - \arctan{v}) + C$$, where C is the constant of integration.
06

Substitute Back

Now, let's substitute back the original variables to get the final result: $$\frac{1}{x^4}(\ln{(\sqrt{u-1})} - \arctan{(\sqrt{u-1})}) + C = \frac{1}{x^4}(\ln{(\sqrt{x^2-1})} - \arctan{(\sqrt{x^2-1})}) + C$$ And that's the final evaluated integral.

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