Integral Of Arctan

w Built-in Arcton (9x) dx?

. "Arctan (9x) dx =

He asserted that his confession had been obtained through torture

The latest difference makes it easier to integrate rational functions

Not really:

dx = dv '(merged on both sides)' x = v

Arctan (9x) = U   '(difference on both sides)  Â' {9 / [1 + (9x) ²] dx = you

This is achieved through partial integration:

ˆ "u dv = u v ˆÂ" v du   '

آر "Arctan (9x) dx = x Arctan (9x) ˆÂ" x {9 / [1 + (9x) ²] dx =

x Arctan (9x)  «[9x / (1 + 81x²)] dx =

Divide the remainder by 18 and multiply to make a number

Derived from the letter, which 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 the ABS value is not necessary because the LN argument is positive)

So the last answer is:

"Arctan (9x) dx = x Arctan (9x) (1/18) LN (1 + 81x²) + C

I can help you ...

Goodbye!

. ct Arctan (9 · x) dx

The application of partial integration is about achieving as much integration and integration as possible. There is only one thing that is not an integral part of it. So it will only connect to dx.

The second use is to see what the difference is in making it easier when it is different. Fortunately, the ultimate antics are easier when they differ:

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

One of these apps will guide you through the app section for proper integration:

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Let's = Arctan (9 · x)

If du / dx = 1 / [(9Â,  · x) Â, ² + 1] Â, · · d / dx [9Â, · · x]

And there is = 9 / (81à· x² + 1) dx

Let's DV = DX

Therefore, "dv = ˆÂÂ" dx

And v = x

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 'u · v ˆÂ "VD

= {Arctan (9 · x) ‚à· {x} ˆÂ "{x} ‚ · {9 / (81 · xò + 1) dx }

= xÃÂ,  ct Arctan (9   · x)  «9    · x / (81    · x² + 1) dx

The source of the contents of a function (power of 1, mutual function) is multiplied by (9%) and differs only with the constant. Change:

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Let's say Q = 81 · xÂò + 1

So dq / dx = 162 × ×

And dx = dq / (162x)

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à'xÃÂ,  · Arctan (9Â,  · x) ˆÂ «9ÂÂ,  · x / (q) DQ / (162àX)

= xÃÂ,  · Arctan (9Â,   x)  «1 / (q) DQ / 18

= xÂà· Arctan (9 · x) (1/18) ‚ÂÈ «1 / (q) DQ

= x · Arctan (9 · x) (1/18) ‚ · ln | q | + C

Reverse alternative to Q:

= x · Arctan (9 · x) (1/18) ‚ · ln | 81Âà· x² + 1 | + C

The content of LN is always positive, so you will definitely not need a value proposition.

= xÂàArctan (9 x) (1/18) ‚ · ln (81ÂàxÂà ² + 1) + C

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Returns:

کٹ «Arctan (9Â,  · x) dx = xÃÂ,  ct Arctan (9Â,  · x) LN (81 x² + 1) / 18 + C

Write as mandatory arcton (9x) x 1

Use u = Arctan (9x) then du / dx = 1/9 x 1 / (1+ (9x) 2) (?)

Use dv / dx = 1, then v = x

So use in the section

Let's U = Arctan (9x), DV = DX.

Integral Of Arctan

Integral Of Arctan

w Built-in Arctan (9x) dx? 3

I know you're using patchwork integration, but that's all I have.

Arctan (9x) dx =

As he emphasizes, integration into parts is necessary, unless

The latest distinction makes it easier to integrate rational functions.

Not really:

dx = dv '(connect both sides)' x = v

Arctan (9x) = u  '(different on both sides)' {9 / [1 + (9x) ²] dx = you

There it is achieved through partial integration:

ˆ "u dv = u v ˆÂ" v du  '

Arctan (9x) dx = x Arctan (9x) ˆ "x {9 / [1 + (9x) ²]} dx =

x Arctan (9x)  «[9x / (1 + 81x²)] dx =

Divide and multiply the remaining integral by 18 to make a number.

The derivative of the denominator, which 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 necessary because ln argument is positive)

So the final answer is:

آر "Arctan (9x) dx = x Arctan (9x) (1/18) ln (1 + 81x²) + C

I can help you ...

Goodbye

Arctan (9 · x) dx

The purpose of applying for partial integration is to achieve maximum integration and integration. There is only one thing that has no integration. So it will only integrate dx.

The second use is to see what makes it easier when it is different and to differentiate between it. Fortunately, inverse elements are simplified when they differ:

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

Here's one of the apps that will guide you through the app sections for proper integration:

"

Let u = Arctan (9ÂÂx)

If du / dx = 1 / [(9Â,  · x) Â, ² + 1] Â,  · d / dx [9Â,  · x]

And is = 9 / (81à· x² + 1) dx

dv = dx two

Therefore, "dv = ˆÂÂ" dx

And v = x

-

Âà 'uÂÃ,  · v ˆÃ,  "v de

= {Arctan (9 · x) ‚ · {x} ˆÂ "{x} ‚ · {9 / (81Âà · xò + 1) dx

= xÃÂ,  · arctan (9Â,   x)  «9   · x / (81   · x² + 1) dx

The derivative of the material of a function (up to the power of 1, reciprocal function) is multiplied by (9%) and differs with only one constant. Change:

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Leave q = 81 · xÂò + 1.

So dq / dx = 162 × ×

And dx = dq / (162 × ×)

-

à'xÃÂ,  · arctan (9Â,  · x) «9ÂÂ,  · x / (q) dq / (162àX)

= 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

Inverse alternative to q:

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

The content of ln is always positive, so you don't need an absolute value bar:

= xÃÂàarctan (9 x) (1/18) ‚ · ln (81 xÂò + 1) + C

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the answer:

Arctan (9Â,  · x) dx = xÃÂ,  · Arctan (9Â,  · x) ln (81 x² + 1) / 18 + C

Write as integral arcton (9x) x 1.

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

Use dv / dx = 1, then v = x

So use in section

Let's say u = arcton (9x), dv = dx.

Integral Of Arctan