# Prove $\int f(x)f'(x)\,dx = \frac {[f(x)]^2}2 + c$ through substitution

I've always taken integration by substitution for granted but recently I've learned that differentials can't fully be treated as variables and that the process of integration by substitution is really just a shorthand. Now, writing something like $$du=u'(x)dx$$ during substitution feels weird and empty.

I understood the justification for $$\int f(g(x))g'(x)\,dx = \int f(u)\,du$$ (given $$u = g(x)$$) from this page, but how do I apply this process to prove $$\int f(x)f'(x)\,dx = \int v\,dv$$ (given $$v=f(x)$$)?

Sorry that I don't have any work to show for myself, it's because I don't even know how to approach this problem.

I'll flip the $$f$$ and $$g$$ around, i.e. lets say that we can assume $$\int f(g(x))g'(x)\,dx = \int f(u)\,du$$ and we want to prove $$\int g(x) g'(x)\, dx = \int u\, du$$ We can write this as $$\int f(g(x))g'(x)\,dx$$ using $$f(x)=x$$, so that the substitution $$u=g(x)$$ gives the result.

Alternatively, since $$(g^2)' = 2gg'$$,$$\int g(x) g'(x)\, dx =\int \left(\frac{g^2}2\right)' \, dx = \frac{g^2}2 + C$$

• Exactly what I wanted, thank you – Typo Oct 25 '18 at 9:58
• (there was a typo in my alternative solution, forgot to bring down the 2) – Calvin Khor Oct 25 '18 at 10:14

This follows from the fundamental theorem of calculus (FTC) and the chain rule.

By the chain rule $$(\frac{f(x)^2}2)'=f(x)\cdot f'(x)$$.

Then by (FTC) $$\int f(x)\cdot f'(x)\operatorname{dx}= \frac{f(x)^2}2+C$$.

To see the last part, write $$g(x)=\frac{f(x)^2}2$$. Then we have $$\int g'(x)\operatorname{dx}=g(x)+C$$.

$$I=\int f(x)f'(x)dx$$ let $$u=f(x)$$. Therefore $$du=f'(x)dx$$. This substitution gives $$I=\int udu$$ $$I=\frac{u^2}2+C$$ $$I=\frac{f(x)^2}{2}+C$$

• I think you mistook my question, the title was not my whole question. What I really wanted to know is how substitution works behind the scenes in proving the integral at hand. Writing "$du = f'(x)dx$" would just be a circular argument – Typo Oct 25 '18 at 19:23