Solve F x = ∫ (from 1 to x) of 1/1+t^3dt

$$F x = \int _ { 1 } ^ { x } \frac { 1 } { 1 + t ^ { 3 } } d t$$

Solve for F

$$F=-\frac{-18\left(\lim_{t\to x}\frac{-\ln(t^{2}-t+1)+2\sqrt{3}\arctan(\frac{\sqrt{3}\left(2t-1\right)}{3})+2\ln(|t+1|)}{6}\right)+\pi \sqrt{3}+6\ln(2)}{18x}$$

$$x\neq 0\text{ and }t\neq -1$$

Steps for Solving Linear Equation

The equation is in standard form.

$$xF=\frac{18\left(\lim_{t\to x}\frac{-\ln(t^{2}-t+1)+2\sqrt{3}\arctan(\frac{\sqrt{3}\left(2t-1\right)}{3})+2\ln(|t+1|)}{6}\right)-\pi \sqrt{3}-6\ln(2)}{18}$$

Divide both sides by x.

$$\frac{xF}{x}=\frac{\left\{\begin{matrix}\frac{-3\ln(x^{2}-x+1)+6\sqrt{3}\arctan(\frac{\sqrt{3}\left(2x-1\right)}{3})+6\ln(|x+1|)-\pi \sqrt{3}-6\ln(2)}{18},&\\-\infty,&\end{matrix}\right.}{x}$$

Dividing by x undoes the multiplication by x.

$$F=\frac{\left\{\begin{matrix}\frac{-3\ln(x^{2}-x+1)+6\sqrt{3}\arctan(\frac{\sqrt{3}\left(2x-1\right)}{3})+6\ln(|x+1|)-\pi \sqrt{3}-6\ln(2)}{18},&\\-\infty,&\end{matrix}\right.}{x}$$

Divide \left\{\begin{matrix}\frac{6\sqrt{3}\arctan(\frac{\sqrt{3}\left(2x-1\right)}{3})-3\ln(-x+1+x^{2})+6\ln(|x+1|)-\pi \sqrt{3}-6\ln(2)}{18},&\\-\infty,&x=-1\end{matrix}\right. by x.

$$F=\left(\left\{\begin{matrix}\frac{-3\ln(x^{2}-x+1)+6\sqrt{3}\arctan(\frac{\sqrt{3}\left(2x-1\right)}{3})+6\ln(|x+1|)-\pi \sqrt{3}-6\ln(2)}{18x},&\\\frac{-\infty}{x},&\end{matrix}\right.\right)$$