Solve the integral: ∫02dxx(x−1)

Question

Solve the integral:  ∫02dxx(x−1)

encolatgehu · Accepted Answer

You can break up the integral at any point or points you like. In this case, you could break it up into three integrals: pick a point c strictly between 0 and 1, and consider:

\int_{0}^{2} \frac{d x}{\sqrt{x} (x - 1)} = \int_{0}^{c} \frac{d x}{\sqrt{x} (x - 1)} + \int_{c}^{1} \frac{d x}{\sqrt{x} (x - 1)} + \int_{1}^{2} \frac{d x}{\sqrt{x} (x - 1)}

= lim_{a \to 0^{+}} \int_{a}^{c} \frac{d x}{\sqrt{x} (x - 1)} + lim_{b \to 1^{-}} \int_{c}^{b} \frac{d x}{\sqrt{x} (x - 1)} + {lim}_{d}^{2} \frac{d x}{\sqrt{x} (x - 1)}

The original improper integral exists if and only if each of the three improper integrals exist.

Jenny Bolton · Accepted Answer

Consider the change of variables t=x. It transforms the interval [0,2] into [0,2] and removes one of the problem zeroes. It is easy to see from the resulting expression that the integral diverges.

karton · Accepted Answer

Try to solve the integral. You&#039;ll find whether it converges. For example, ∫dxx(1−x) with ϵ&amp;gt;0. Later on, you can study the limit ϵ→0+. In this way, you &#039;&#039;kill two birds with a one shot&#039;&#039;.

Solve the integral: \int_0^2\frac{dx}{\sqrt{x}(x-1)}

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