91Ó°ÊÓ

The given integrand is \(\frac{1}{x\sqrt{x^2-100}}\). We can recognize that it is in the form of a hyperbolic trigonometric function. This suggests that we can use the following substitution to simplify the integrand: $$x = 10\cosh(t)$$ Now we need to find the derivative \(dx\) to perform the substitution.

Taking the derivative on both sides, we get: $$\frac{d x}{d t} = 10\sinh(t)$$ Multiplying both sides by dt, we have: $$dx = 10\sinh(t)\, dt$$ Now, we are ready to substitute \(x\) and \(dx\) in the given integral.

Substituting \(x = 10\cosh(t)\) and \(dx = 10\sinh(t)\, dt\) in the given integral, we have: $$\int \frac{1}{x\sqrt{x^2-100}}\, dx = \int \frac{10\sinh(t)\, dt}{10\cosh(t)\sqrt{(10\cosh(t))^2-100}}$$ Now, we need to simplify the integrand.

First, we notice that \(10\) can be factored out from the integrand, which gives us: $$\int \frac{10\sinh(t)\, dt}{10\cosh(t)\sqrt{(10\cosh(t))^2-100}} = 10\int \frac{\sinh(t)\, dt}{\cosh(t)\sqrt{(10\cosh(t))^2-100}}$$ Next, let's simplify the denominator by using the properties of hyperbolic trigonometric functions: $$\cosh^2(t)-1=\sinh^2(t)$$ Now, $$\begin{aligned} (10\cosh(t))^2-100 &= 100\cosh^2(t)-100 \\ &= 100(\cosh^2(t)-1) \\ &= 100\sinh^2(t)\end{aligned}$$ Plugging this back to the integrand, we have: $$10\int \frac{\sinh(t)\, dt}{\cosh(t)\sqrt{100\sinh^2(t)}} = \int \frac{10\sinh(t)\, dt}{\sinh(t)}$$ Now, we can simplify further since \(\sinh(t)\) cancels out: $$\int \frac{10\sinh(t)\, dt}{\sinh(t)} = \int 10\, dt$$

Now that we have a simpler integrand, we can easily find the antiderivative: $$\int 10\, dt = 10t + C$$ We have found the antiderivative in terms of \(t\), so we need to substitute back \(x\) for \(t\).

Recalling our substitution \(x = 10\cosh(t)\), we can rewrite it as: $$t = \cosh^{-1}\left(\frac{x}{10}\right)$$ Now, substituting this back in our antiderivative, we get: $$\int\frac{1}{x\sqrt{x^2-100}}\,dx = 10\cosh^{-1}\left(\frac{x}{10}\right) + C$$

Differentiating our result with respect to x, we have: $$\frac{d}{dx}\left(10\cosh^{-1}\left(\frac{x}{10}\right) + C\right) = \frac{1}{\sqrt{(\frac{x}{10})^2-1}} \cdot 10\cdot \frac{1}{x} = \frac{1}{x\sqrt{x^2-100}}$$ This differentiation result confirms the correctness of our determined indefinite integral.