# Antiderivative of arctan and x function (by parts maybe? )

antiderivative of arctan and x function (by parts... maybe??!!)

find antiderivative of 2xarctan(8x)

{ = antiderivative s thing.

I did it by parts at first and got

Code:
x^2arctan(8x) - 8 * { (x^2)/(1+64x^2)

To get the antiderivative of the second part i did it by parts again and that ended up bringing me back to the same equation that i first derived.
Im pretty lost here so any help would be really appreciated!!!
Thanks!

Assuming I'm understanding your work correctly:

Original integral = Blah minus Original Integral.

Correct?

If so, just isolate the original integral by adding it to both sides and divide by two. Simple algebra hidden in calculus. ;)

Edit: Don't forget to throw a + C in there too, since it is an indefinite integral.

Edit Numero Two: I just went through the problem, and I had initially thought it was harder than it is. Just make u = 1+64x^2 in the second integral and you should have no problem.

Last edited:
You're right that you need to use integration by parts and you're going about it correctly as far as I can tell. You just need to take the integral of your v*du and add your constant.

alright so gave er a shot (or a hundred).
Heres where im at :

After first integration by parts:
x^2arctan(8x) - 8 * { (x^2)/(1+64x^2)

Now only the second part integration by parts:
-8 * { (x^2)/(1+64x^2)
dv = 1 / (64x^2+1) u = x^2
v = arctan(8x)/8 du= 2x

-8 * [ x^2*arctan(8x)/8 - { 2x*arctan(8x)/8 ]
= -x^2*arctan(8x) + { 2x*arctan(8x)

Note: To find the solution add back in the x^2arctan(8x) of the equation in the code box of my first post (took it out just to find the antiverivative of second half).

this yields:
x^2arctan(8x) - x^2*arctan(8x) + { 2x*arctan(8x)
= { 2x*arctan(8x)
which is the original problem.

Sorry for being such a hassell! Thanks for your help i really appreciate it!

$$- \int \frac{8x^2}{64x^2 +1}dx$$

you should be able to simplify that without using integration by parts again, any ideas?

Unfortunately, in this case, Cemar, that results in 0 = 0, which does not help you much. Like I edited above, just use the LIPET rule for u substitutions (logarithms, inverse trig, polynomials, exponentials, trig--in that order) to decide on a u value and do a simple integration.

I came back and looked at this after finishing my meal (I had been multitasking earlier), and I noticed that I made a mistake in my work before. :P

Like RyanSchw was saying, just evaluate that integral (I personally used partial fractions, though I'm sure there may be other methods) and you should be good to go!

Thanks!! you guys are awesome!!!!!
=)