Question

What is the integral of `int tan^-1(1/x)dx`?

Answer 1

`xcot^-1(x)+1/2ln(1+x^2)+C`

Explanation:

You may notice that `tan^-1(1/x)=cot^-1(x)`.

If this isn't natural to you, here's why:

`A=tan^-1(1/x)=>tan(A)=1/x=>cot(A)=x=>cot^-1(x)=A`

So, we see that:

`inttan^-1(1/x)dx=intcot^-1(x)dx`

From here, use integration by parts, which states that:

`intudv=uv-intvdu`

So, let `u=cot^-1(x)=>du=-1/(1+x^2)dx` and
`dv=dx=>v=x`.

Thus:

`intcot^-1(x)dx=xcot^-1(x)-int(-x)/(1+x^2)dx`

Use substitution for the remaining integral (you may notice this is shaping up to resemble the natural log integral...).

Let `t=1+x^2` so that `dt=2xdx`.

`intcot^-1(x)dx=xcot^-1(x)+1/2int(2x)/(1+x^2)dx`

`=xcot^-1(x)+1/2int(dt)/t`

`=xcot^-1(x)+1/2ln(abst)+C`

`=xcot^-1(x)+1/2ln(1+x^2)+C`

Answer 2

I got:

`xarctan(1/x) - ln|1/x| + 1/2ln|1/x^2 + 1| + C`

If you merge the `ln` terms with some logarithmic tricks, you might also write this more concisely as:

`xarctan(1/x) + 1/2ln(1 + x^2) + C`

---

For this, you kind of have to experiment. I end up doing a few substitutions, one integration by parts, and some partial fractions, but you may find another way.

(You can probably start with integration by parts, and use the Chain Rule.)

For our first substitution, let `u = 1/x`. Then, `du = -1/(x^2)dx`, and `dx = -1/u^2du`:

`= -intarctanu/(u^2)du`

`= int -1/u^2 arctanudu`

Note how we manipulated our substitution to basically incorporate an additional term into the original integrand; we got `arctan(1/x) = -1/u^2arctanu`.

Now, we can use integration by parts, because we should know that `d/(dx)[arctanx] = 1/(1+x^2)`, and `int -1/x^2dx = 1/x + C`.

Therefore, for

`\mathbf(int sdt = st - inttds)`,

Let

`s = arctanu,`
`dt = -1/u^2du,`
`t = 1/u,`
`ds = 1/(1+u^2)du`.

Now what you have is:

`= arctanu/u - int1/(u(1+u^2))du`

For the integral here, you'll have to use partial fractions. For now let's focus on the integral without the subtraction sign.

`int1/(u(1+u^2))du = A/u + (Bu + C)/(u^2 + 1)`

Ignoring the integral symbols for now:

`=> [A(u^2 + 1) + (Bu^2 + Cu)]/(u(u^2 + 1))`

`= [Au^2 + A + Bu^2 + Cu]/(u(u^2 + 1))`

`= [(A + B)u^2 + (C)u + A]/cancel((u(u^2 + 1))) = 1/cancel((u(u^2+1)))`

So, you have the comparisons of each term in the quadratic expression.

Writing `color(highlight)((A + B))u^2 + color(red)((C))u + color(orange)(A) = color(highlight)(0)x^2 + color(red)(0)x + color(orange)(1)`, we get:

`A + B = 0`
`C = 0`
`A = 1`

Therefore, it's a simple solution: solve the system, and you get

`A = 1`,
`B = -1`,
`C = 0`.

This results in the integral becoming:

`int1/(u(1+u^2))du = int1/udu - int u/(u^2 + 1)du`

`= ln|u| - 1/2int (2udu)/(u^2 + 1)`

One more substitution. Let `z = u^2 + 1`, so `dz = 2udu`, giving you:

`= ln|u| - 1/2 int 1/zdz`

`= color(green)(ln|u| - 1/2 ln|u^2 + 1|)`

Combine this result back with our initial integral to get:

`=> arctanu/u - [ln|u| - 1/2 ln|u^2 + 1|]`

`= arctanu/u - ln|u| + 1/2 ln|u^2 + 1|`

Now, our original substitution was `u = 1/x`, so we'd get:

`= xarctan(1/x) - ln|1/x| + 1/2ln|1/x^2 + 1| + C`

@mason-m got a seemingly different answer, but let's manipulate this answer to see how it's actually the same.

`=> xarctan(1/x) + 1/2*2*ln|x| + 1/2ln|1/x^2 + 1| + C`

`= xarctan(1/x) + 1/2ln|x|^2 + 1/2ln|1/x^2 + 1| + C`

`= xarctan(1/x) + 1/2ln(x^2) + 1/2ln(1/x^2 + 1) + C`

`= xarctan(1/x) + 1/2ln(x^2(1/x^2 + 1) + C`

`= color(blue)(xarctan(1/x) + 1/2ln(1 + x^2) + C)`

(which equals `x"arc" cotx + 1/2ln(1 + x^2) + C` as well from @mason-m's answer.)

Answer 3

See the Note in Explanation.

Explanation:

I have to just add a little to the Answers submitted by Turong - Son N. and mason m :

`tan^-1 (1/x)=cot^-1 x, if x>0`

But, the case, `if x<0, tan^-1 (1/x)=cot^-1x-pi`, needs to be addressed.

In this case, `inttan^-1 (1/x)dx=int(cot^-1 x-pi)dx`

`=I-pix+C`, where `I` is the integral readily derived by either of the contributors.