the integral:
$$\int_0^1 \frac{1}{x^2} = lim_{t \to o^+} \int_0^1 \frac{1}{x^2}= lim_{t \to o^+} \frac{1}{x} \vert_t^1= lim_{t \to o^+} (-1+\frac{1}{t})$$
I dont understand how we go from $x^2$ to $x$
2026-04-06 04:54:57.1775451297
question regarding the solution process of an improper integral
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The steps are wrong. It should be
$$\int_0^1 \frac{1}{x^2}\, dx = \lim_{t\to 0^+} \int_t^1 \frac{1}{x^2}\, dx = \lim_{t\to 0^+} -\frac{1}{x}\Bigl|_t^1 = \lim_{t \to 0^+} \left(-1 + \frac{1}{t}\right)$$ The second equality is obtained using the fact that $-\frac{1}{x}$ is an antiderivative of $\frac{1}{x^2}$.