Find \(\int \frac{\mathrm{d} x}{(1-x) \sqrt{x}}\).

Short Answer

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Question: Evaluate the integral \(\int \frac{\mathrm{d} x}{(1-x) \sqrt{x}}\). Answer: \(\int \frac{\mathrm{d}x}{(1-x) \sqrt{x}} = \frac{1}{2}\ln\left|\frac{1+\sqrt{x}}{1-\sqrt{x}}\right| + C\)

Step by step solution

01

Choose an appropriate substitution

Let's make a substitution to simplify the integrand. We will choose \(u = \sqrt{x}\), so \(x = u^2\). To find the differential, we take the derivative of \(x\) with respect to \(u\) and get \(\mathrm{d}x = 2u\,\mathrm{d}u\).
02

Substitute and simplify integrand

Now, let's substitute the expressions for \(x\) and \(\mathrm{d}x\) into the original integral: $$\int \frac{\mathrm{d}x}{(1-x)\sqrt{x}} = \int \frac{2u\,\mathrm{d}u}{(1-u^2)u}.$$ Simplifying the integrand, we get: $$\int \frac{2u\,\mathrm{d}u}{(1-u^2)u} = 2\int \frac{\mathrm{d}u}{1-u^2}.$$
03

Recognize a known integral

The simplified integral is now in the form of a known integration result: $$\int \frac{\mathrm{d}u}{1-u^2}.$$ This integral can be found using the definition of the natural logarithm function and the properties of hyperbolic functions. The result is: $$\int \frac{\mathrm{d}u}{1-u^2} = \frac{1}{2}\ln\left|\frac{1+u}{1-u}\right| + C.$$
04

Substitute back for x

Finally, let's substitute back in terms of \(x\) using the original substitution \(u = \sqrt{x}\): $$\frac{1}{2}\ln\left|\frac{1+\sqrt{x}}{1-\sqrt{x}}\right| + C.$$ So, the antiderivative of the given integral is: $$\int \frac{\mathrm{d}x}{(1-x) \sqrt{x}} = \frac{1}{2}\ln\left|\frac{1+\sqrt{x}}{1-\sqrt{x}}\right| + C.$$

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