Problem involving Fundamental Theorem of Calculus: simplifying $\frac{\mathrm d}{\mathrm dx}\int_x^{x^2}\frac{t}{\log t}\,\mathrm dt$

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I'm working on the following question:

Simplify the following:
$$\frac{\mathrm d}{\mathrm dx}\int_x^{x^2}\frac{t}{\log t}\,\mathrm dt$$

The solution key says this simplifies to:
$$2x\frac{x^2}{\log(x^2)}-\frac{x}{\log(x)}$$

I think this wrong though. For two reasons:

  1. We can't just use fundamental theorem of calculus (FTC) because the integrand needs to be continuous on the interval of interest. The quotient of continuous functions is continuous, provided the denominator is non-zero. There's a problem though at $t=1$.
  2. FTC is stated with one end fixed — can we just assume both ends are functions?

Given these caveats, I'm not sure how to proceed — thoughts?

Best Answer

The domain of the function $f(x) = \displaystyle \int_x^{x^2} \dfrac{t}{\log t} \, dt$ is $(0,1) \cup (1,\infty)$, but it would certainly be consistent to define $f(1) = 0$ since that integral is defined on a degenerate interval. Take the domain to be $(0,1) \cup (1,\infty)$.

Now, let $F$ be an antiderivative of $\dfrac{t}{\log t}$ defined on either interval $(0,1)$ or $(1,\infty)$. The fundamental theorem of calculus gives you $$ f(x) = F(x^2) - F(x)$$ so that $$f'(x) = F'(x^2) \cdot 2x - F'(x) = \frac{x^2}{\log (x^2)} \cdot 2x - \frac{x}{\log x}.$$

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