How do you integrate arctan(9x) dx?

Question

i know you use integration by parts but that was as far as i got.

Answer 1

∫ arctan( 9·x ) dx

One approach to integration by parts is to take the biggest part that you can integrate and integrate it. Here, there is just one thing that doesn't have an integral. So you'll just integrate the dx.

Another approach is to take something that simplifies if it is differentiated and differentiate it. Fortunately, inverse tangent simplifies if it is differentiated:

d/dx[ arctan(u) ] = 1/(u² + 1) · du/dx

Either of these approaches here would bring you to the proper integration by parts approach:

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Let u = arctan( 9·x )

Then du/dx = 1/[ ( 9·x )² + 1 ] · d/dx [ 9·x ]

And du = 9/( 81·x² + 1 ) dx

Let dv = dx

Then ∫dv = ∫dx

And v = x

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→ u·v - ∫ v du

= { arctan( 9·x ) } · { x } - ∫ { x } · { 9/( 81·x² + 1 ) dx }

= x·arctan( 9·x ) - ∫ 9·x/( 81·x² + 1 ) dx

The derivative of the contents of one function (the power of -1, the reciprocal function) is multiplied outside of it (the 9·x), varying only by a constant. Make a substitution:

————————————————

Let q = 81·x² + 1

Then dq/dx = 162·x

And dx = dq/( 162·x )

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→ x·arctan( 9·x ) - ∫ 9·x/( q ) dq/( 162·x )

= x·arctan( 9·x ) - ∫ 1/( q ) dq/18

= x·arctan( 9·x ) - (1/18) · ∫ 1/( q ) dq

= x·arctan( 9·x ) - (1/18) · ln| q | + C

Reverse substitute for q:

= x·arctan( 9·x ) - (1/18) · ln| 81·x² + 1 | + C

The contents of ln is always positive so you don't need absolute value bars:

= x·arctan( 9·x ) - (1/18) · ln( 81·x² + 1 ) + C

————————————————

Answer:

∫ arctan( 9·x ) dx = x·arctan( 9·x ) - ln( 81·x² + 1 ) / 18 + C

Answer 2

write it as the integral of arctan(9x) x 1

use u=arctan(9x) then du/dx= 1/9 x 1/(1+(9x)^2)       (?)

use dv/dx = 1 then v =x

then use by parts

Answer 3

Let u=arctan(9x), dv=dx.

du=(1/(1+(9x)^2)) <----go back to your trigonometric integration lessons for why.

int(dv)=int(dx), v=x

int(arctan(9x)dx)=(x)(arctan(9x))-int (x/(1+(9x)^2))dx)    +c lol

At least I think that's it, please check it before accepting it, I'm a bit rusty.

Answer 4

∫ arctan(9x) dx =

as you correctly point out, integration by parts is appropriate, in that

arctangent differentiation yields a rational function which is easier to integrate;

thus let:

dx = dv → (integrate both sides) → x = v

arctan(9x) = u → (differentiate both sides) → {9 /[1 + (9x)²]} dx = du

therefore, integrating by parts, you get::

∫ u dv = u v - ∫ v du →

∫ arctan(9x) dx = x arctan(9x) - ∫ x {9 /[1 + (9x)²]} dx =

x arctan(9x) - ∫ [9x /(1 + 81x²)] dx =

divide and multiply the remaining integral by 18, so that the numerator turns into

the derivative of the denominator, that is 162x:

x arctan(9x) - (1/18) ∫ [(18)9x /(1 + 81x²)] dx =

x arctan(9x) - (1/18) ∫ [162x /(1 + 81x²)] dx =

x arctan(9x) - (1/18) ∫ [d(1 + 81x²)]/ (1 + 81x²) =

x arctan(9x) - (1/18) ln (1 + 81x²) + C

(note that abs value is not needed, since ln argument is positive)

thus, the definitive answer is:

∫ arctan(9x) dx = x arctan(9x) - (1/18) ln (1 + 81x²) + C

I hope it helps...

Bye!