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

[cemar.]

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

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!

 

[Alienjoey]

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.

 

[RyanSchw]

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.

 

[cemar.]

alright so gave er a shot (or a hundred).
Heres where I am 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!

 

[RyanSchw]

When your integral becomes

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

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

 

[Alienjoey]

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.

 

[Alienjoey]

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!

 

[cemar.]

Thanks! you guys are awesome!
=)