Question

integration by parts, please help :)

Answer 1

\[\int\limits_{}^{}3Arctan(2x)dx\]

Answer 2

so I did u= tan^-1(2x), so then du= 1/2+2x^2 and I did dv= 3 dx, so then v = 3x

Answer 3

\[\Large u= \tan^{-1}(2x) \\
\Large \frac{du}{dx}=\frac{2}{4x^2+1} \]
Yes?

Answer 4

oh, good call I did that derivative wrong

Answer 5

actually, wait, sorry, where did the 4 come from? because the derivative of tan^-1 is 1/(1+x^2) isn't it?

Answer 6

So you want to achieve the following \[\Large \int3 \tan^{-1}(2x)dx= 3x\tan^{-1}(2x)- \int 3x \cdot \frac{4}{4x^2+1}dx \]

Answer 7

\[ \Large (\tan(ax))' = \frac{a}{a^2x^2+1} \]

Answer 8

I am missing the ^(-1) in the above of course, I am talking about the arctan function, not about the tan function

Answer 9

I thought with the chain rule you took the derivative of arctan, which is 1/(1+x^2) and then multiplied by 2 because of the chain rule. (?)

Answer 10

sorry, I'm just a little confused about that for some reason

Answer 11

oh no wait, I see it now, sorry.

Answer 12

What you're right about is that my integrand above is faulty, it is supposed to be a -> 2 in the numerator.

Answer 13

I was forgetting to plug in the 2x in place of x. okay, let me see if I can solve this now, just a sec

Answer 14

thanks for the help by the way

Answer 15

no problem.

Answer 16

Okay, I ended up with the right answer. Thanks :)

Answer 17

\[\Huge \checkmark \] Well done