Integral of $\int \tan(x) \sec^3 {x}\, \text dx$

You missed a $-$ sign in the denominator. $$\int x^{n} \ \text dx = \frac{x^{n+1}}{n+1} + C$$ when $n \neq -1$.

So when $n=-4$ you get $$\int x^{-4} \ \text dx = -\frac{1}{3} \cdot x^{-3} +C$$

There is still an easier method of doing this: $$\int \tan{x}\cdot \sec^{3}(x) \ \text dx = \int \tan{x} \cdot \sec{x} \cdot \sec^{2}(x) \ \text dx = \int t^{2} \ \text dt$$ by putting $t=\sec{x}$.