Resolver ejercicio 4 integrales teorema fundamental calculo

La integral que planteas, ya te la resolví en otro ejercicio, ahora veamos la derivada del resultado...

$$\begin{align}&\int_a^x \frac{1}{t^3+1}dt = \int_a^x \frac{2-t}{3(t^2-t+1)}dt + \int_a^x \frac{1}{3(t+1)}dt =\\&1/6 \bigg(-ln (t^2-t+1)+2ln(t+1)+2 \sqrt{3}arctan \bigg(\frac{2t-1}{\sqrt 3}\bigg) \bigg) \Bigg|_a^x=\\&1/6 \bigg(-ln (x^2-x+1)+2ln(x+1)+2 \sqrt{3}arctan \bigg(\frac{2x-1}{\sqrt 3}\bigg) \bigg) - 1/6 \bigg(-ln (a^2-a+1)+2ln(a+1)+2 \sqrt{3}arctan \bigg(\frac{2a-1}{\sqrt 3}\bigg) \bigg) \\&\text{La derivada respecto a "x", sería (recordá que "a" es una constante)}\\&\frac{d}{dx} \Bigg(\int_a^x \frac{1}{t^3+1}dt \Bigg)=\\&\frac{d}{dx} \Bigg( 1/6 \bigg(-ln (x^2-x+1)+2ln(x+1)+2 \sqrt{3}arctan \bigg(\frac{2x-1}{\sqrt 3}\bigg) \bigg) - 1/6 \bigg(-ln (a^2-a+1)+2ln(a+1)+2 \sqrt{3}arctan \bigg(\frac{2a-1}{\sqrt 3}\bigg) \bigg) \bigg) =\\&\frac{d}{dx} \Bigg( 1/6 \bigg(-ln (x^2-x+1)+2ln(x+1)+2 \sqrt{3}arctan \bigg(\frac{2x-1}{\sqrt 3}\bigg) \bigg)\bigg) - \frac{d}{dx} \Bigg(1/6 \bigg(-ln (a^2-a+1)+2ln(a+1)+2 \sqrt{3}arctan \bigg(\frac{2a-1}{\sqrt 3}\bigg) \bigg) \bigg) =\\&\frac{d}{dx} \Bigg( 1/6 \bigg(-ln (x^2-x+1)+2ln(x+1)+2 \sqrt{3}arctan \bigg(\frac{2x-1}{\sqrt 3}\bigg) \bigg)\bigg) - 0=\\&\frac{d}{dx} \Bigg( 1/6 \bigg(-ln (x^2-x+1)\Bigg)+\frac{d}{dx} \Bigg(2ln(x+1)\Bigg)+\frac{d}{dx} \Bigg(2 \sqrt{3}arctan \bigg(\frac{2x-1}{\sqrt 3}\bigg) \bigg)\bigg)=\\&\frac{-1(2x-1)}{6(x^2-x+1)} +\frac{2}{x+1} +2 \sqrt{3}  \frac{1}{1+(\frac{2x-1}{\sqrt 3})^2}2=\\&\frac{-2x+1}{6(x^2-x+1)} +\frac{2}{x+1} +  \frac{4 \sqrt{3}}{1+(\frac{2x-1}{\sqrt 3})^2}\end{align}$$