Integration By Parts

Integration By Parts - this is a method for solving integrals involving products of functions. By products of functions we mean:

f(x)g(x), or xCos[x], where x is a function and Cos[x] is a function. Here is the parts formula:

$\int_a^b f(x) g'(x)\, dx = \left[ f(x) g(x) \right]_{a}^{b} - \int_a^b f'(x) g(x)\, dx\!$

Now, you might wonder how this was derived...well, this is derived from the product rule. Recall the product rule for differentiation:

$(f\cdot g)'=f'\cdot g+f\cdot g' \,$

If this isn't making sense yet don't worry. What you need to realize is that the integral of the left side of the above equation, is simply f*g. Remember that the integral of a derivative, is just a function. Quickly:

f(x) = x^2
f '(x) = 2x
Integral of 2x = (2 *x^2)/2, and the 2s cancel leaving us with x^2 or f(x).

So, since the left and right side of any equation are equivalent :-), we can also integrate the right side of the equation. After recalling the fact above, that the derivative of an integral is just the function and a little rearranging we arrive at the integration by parts formula. Look at the below.

First is the product rule
Then you integrate the derivative to get back the original f(x)*g(x) function.
Then you rearrange the equation...that's it.

(note, in the first line, that's g ' (x), ignore the carrot, there is no exponent)

Let's do a integration by parts problem:

Notice here, U substitution doesn’t work. There’s no trig substitution to fall back on and there is no immediately obvious antiderivative.

Let’s use integration by parts, note, we will be replacing f and g with u and v, don’t get too excited!

$\int u\, dv=uv-\int v\, du.\!$

So how do you choose u and v? Good question, let’s think about it for a second. Looking at the integration by parts formula, you have to take the antiderivative of u on the right side of the equation. The goal is always to get a simpler integral on the right side of the equation. So if differentiating u creates a integral more complicated than the original integral on the left, don’t use it.

Note that we choose u, lnx so that it's derivative is easy to integrate, it happens to cancel nicely with the antiderviate of dv, so enjoy that while you can.

So we have our answer, the indefinite integral is highlighted. Note, a few principles used in the simplification included reducing the power of the exponent and eliminating the x in the denominator, and bringing out a constant in front of the integral.

Also - in case that wasn't enough (I doubt it), some great videos on integration by parts on YouTube. And if you like cheap books, check out th

e iChapters ad on the right. You can buy individual chapters, in PDF format for a few bucks a piece...that's a cool deal!