Integration by parts theorem proof.
Matthew Barrera 
Theorem statement:
Let $u(x)$ and $v(x)$ be differentiable functions on some interval $I$ and let $u'(x)v(x)$ have a primitive function on interval $I$, then function $u'(x)v(x)$ also has a primitive function on interval I and the following statement is true:
$$\int u(x)v'(x)dx=u(x)v(x)- \int v(x)u'(x)dx$$
Theorem proof:
First, from the theorem statement I can see that I have to use the fact that $u(x)$ and $v(x)$ are differentiable and that $u'(x)v(x)$ has a primitive function to prove that $u'(x)v(x)$ also has a primitive function on same that interval and that $\int u(x)v'(x)dx=u(x)v(x)- \int v(x)u'(x)dx$ is true on interval $I$.
Now, since $u(x)$ and $v(x)$ are differentiable, their product is differentiable too, so I have:
$$[u(x)v(x)]'=u'(x)v(x)+v'(x)u(x) \Rightarrow u'(x)v(x) = [u(x)v(x)]'-v'(x)u(x)$$
I believe that at this moment I can prove that $u'(x)v(x)$ has primitive function on $I$, since, by the theorem statement $v'(x)u(x)$ has primitive function it means that I only have to prove that $[u(x)v(x)]'$ has a primitive function in order to prove that $u'(x)v(x)$ has a primitive function. I mean, it seems to me that primitive function of $[u(x)v(x)]'$ is $u(x)v(x)$ because if we integrate it we have:
$\int [u(x)v(x)]'dx = \int \frac{d}{dx}u(x)v(x)dx = u(x)v(x) $ which means that $[u(x)v(x)]'$ has primitive function too, so $u'(x)v(x)$ has primitive function, which is what I had to prove. I just have one question here. Is this the correct way of proving this?
Now, the rest of it is easy:
$u'(x)v(x) = [u(x)v(x)]'-v'(x)u(x)$
if we use antiderivative on both sides we get the wanted expression
$$\int u(x)v'(x)dx=u(x)v(x)- \int v(x)u'(x)dx$$
But, the main question is: Is my approach for proving that $u'(x)v(x)$ has primitive function valid?
$\endgroup$1 Answer
$\begingroup$Yes your approach is valid. There was a misprint in your question I correct it.
$\endgroup$ 0
