Evaluate $\int{\frac{1}{x^3}}$ using u-sub.

Question

I tried solving $\int{\frac{1}{x^3}}$ using u-sub instead of power rule and I got $-\frac{1}{2}x^{-2}$ instead of $\frac{x^4}{4}$. Its very possible I've made a very simple mistake or their is something fundamentally wrong with my idea, if someone could show how/if this problem can be done with u-sub it would be greatly appreciated. My work to solve the problem are as follows: $$ \int{\frac{1}{x^3}dx} \\u=x^3 \\x=u^{1/3} \\du=3x^{2}dx \\dx=\frac{1}{3x^2}du \\\int{\frac{1}{u}\times{\frac{1}{3x^2}du}} \\\int{\frac{1}{u}\times{\frac{1}{3u^{2/3}}du}} \\\int{u^{-1}\times{\frac{1}{3}\times{u^{-2/3}}}du} \\\frac{1}{3}\int{u^{-1}\times{{u^{-2/3}}}du} \\\frac{1}{3}\int{u^{-1+-2/3}du} \\\frac{1}{3}\int{u^{-5/3}du} \\\frac{1}{3}\times{\frac{u^{\frac{-5}{3}+1}}{\frac{-5}{3}+1}} \\\frac{1}{3}\times{\frac{u^{\frac{-2}{3}}}{\frac{-2}{3}}} \\\frac{1}{3}\times{\frac{3}{-2}}\times{u^{\frac{-2}{3}}} \\-\frac{1}{2}\times{u^{\frac{-2}{3}}} \\-\frac{1}{2}\times{x^{3^{\frac{-2}{3}}}} \\-\frac{1}{2}\times{x^{\frac{-6}{2}}} \\-\frac{1}{2}x^{-2} $$

Answer 1

Remember that $$\frac{1}{x^3} = x^{-3}$$ So actually by the power rule you'd have:

$$\int x^{-3}dx = \frac{x^{-3+1}}{-3+1}+C = -\frac{1}{2}x^{-2}+C$$

So you are correct with your solution

Answer 2

Your answer is fine maybe there is a typo in your textbook.

Also an easier u-sub would be letting $u=\frac{1}{x} $ then $du=-\frac{1}{x^2} dx $

and then notice $\frac{1} {x^3} = \frac{1}{x^2} \cdot \frac{1}{x}$

So the integral becomes $\int (-u) du = -\frac{u^2}{2}+C $

and after subbing $u $ back you get $-\frac{1}{2} \cdot \frac{1}{x^2}= -\frac{1}{2}\cdot x^{-2}$