# I Integration by Parts without using u, v

1. Nov 30, 2016

### Sang Ho Lee

Hello, I'm currently taking calc 1 as an undergraduate student, and my professor just showed us a new? way of solving Integration By Parts.

This is the example he gave"

Is there a name for this technique that substitutes d(___) instead of dx?

Thank you,

2. Dec 1, 2016

### Staff: Mentor

It doesn't substitute d(___) instead of dx. What you're calling d(___) is just dv.

This is nothing more than integration by parts, but without stating dv explicitly. Here $u = \ln x$ and $dv = xdx$. From the latter, we get $v = \frac{x^2}2$.

3. Dec 1, 2016

### PeroK

I've never used $u, v$ and have never really understood the point. I would rename integration by parts as the "reverse product rule", because it's just the product rule for differentiation in reverse. Anyway, what I do is:

$\int f(x)g(x) dx = f(x) \int g(x)dx - \int f'(x) (\int g(x)dx)dx$ (*)

I don't see why you need any other variables. You pick $f$ because it's easy to differentiate and/or $g$ because you know how to integrate it already. Your one I would do:

$\int (ln(x)) x dx = (ln(x)) \frac{x^2}{2} - \int (\frac{1}{x})(\frac{x^2}{2})dx = \frac{x^2ln(x)}{2} - \int \frac{x}{2} dx = \frac{x^2ln(x)}{2} - \frac{x^2}{4} + C$

I knew how to differentiate $ln(x)$ and how to integrate $x$.

PS To prove (*), I would imagine that you already know the integral of $g$ is $h$, which means $h'(x) = g(x)$. So:

$\int f(x)g(x) + f'(x)h(x)dx = \int f(x)h'(x) + f'(x)h(x)dx = \int \frac{d}{dx} (f(x)h(x))dx = f(x)h(x) + C$

Hence:

$\int f(x)g(x)dx = f(x)h(x) -\int f'(x)h(x)dx+ C$

Where $\int g(x)dx = h(x) + C$

And that seems to me to take some of the "magic" out of integration by parts.