Computing derivative of $\int_0^x \dfrac{1}{t^3 + 1} dt$

Question

I have been asked to find the derivative of this function:

$f(x) = \int_0^x \dfrac{1}{t^3 + 1} d\mathrm{t}$

So far the book I'm working on has covered till the first and second fundamental theorem of calculus.

I tried evaluating the indefinite integrals using Riemann's n-sum but wasn't able to reduce it further. Another attempt by me was using the second fundamental theorem:

$\int_0^x f'(t) \mathrm{d}t = f(x) - f(0)$

But I'm not sure how to proceed from here.

Answer 1

The fundamental theorem of calculus states that if $f$ is continuous on $[a,b]$, then $$ \frac{d}{dx}\int_{a}^{x}f(t) \, dt = f(x) $$ for all $x\in[a,b]$. Let $\displaystyle{f(t)=\frac{1}{t^3+1}}$ and set $a=0$. Since $f$ is a rational function, it is continuous on its domain, and so it is continuous on the nonnegative reals. This means that for any $b>0$, $$ \frac{d}{dx}\int_{0}^{x}\frac{1}{t^3+1} \, dt = \frac{1}{x^3+1} $$ for all $x\in[0,b]$. Since $b$ can be made arbitrarily large, $$ \frac{d}{dx}\int_{0}^{x}\frac{1}{t^3+1} \, dt=\frac{1}{x^3+1} $$ for all $x\in[0,\infty)$.

Answer 2

Deafferenting the second fundamental theorem you've given gives

$\frac{d}{dx}\int_0^x f^\prime (t)dt = f^\prime (x)$

as $f(0)$ does not change with x.

Letting

$f^\prime(t) = \frac{1}{t^3 + 1}$

then gives

$\frac{d}{dx}\int_0^x \frac{1}{t^3 + 1}dt = \frac{1}{x^3 + 1}$