Question

How do you find `int arctanx`?

Answer 1

The answer is `=xarctanx-1/2ln(1+x^2)+C`

Explanation:

Perform this integral by integration by parts

`intuv'=uv-intu'v`

`u=arctanx`, `=>`, `u'=1/(1+x^2)`

`v'=1`, `=>`, `v=x`

Therefore,

The integral is

`intarctanxdx=xarctanx-int(xdx)/(1+x^2)`

`int(xdx)/(1+x^2)=1/2int(2xdx)/(1+x^2)=1/2ln(1+x^2)`

And finally,

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

Answer 2

`xarc tanx-1/2ln(x^2+1)+C, or,`

`xarc tanx-lnsqrt(x^2+1)+C`.

Explanation:

Suppose that, `I=intarc tanxdx=int(arc tanx)(1)dx`

Using the following Rule of Integration by Parts (IBP) :

IBP : `intuv'dx=uv-intu'vdx`.

We take, `u=arc tanx, and, v'=1`.

`:. u'=1/(x^2+1), and, v=intv'dx=int1dx=x`.

`:. I=xarc tanx-intx/(x^2+1)dx`,

`=xarc tanx-1/2int(2x)/(x^2+1)dx`,

`=xarctan x-1/2int{d/dx(x^2+1)}/(x^2+1)dx`.

Since, `int(f'(x))/f(x)dx=ln|f(x)`,

`:. I=xarc tanx-1/2ln(x^2+1)+C, or,`

`I=xarc tanx-lnsqrt(x^2+1)+C`.#.

Answer 3

We use the Integral of Inverse Functions theorem.

Explanation:

If `f^-1` denotes a continuous inverse function then

`\int f^{-1}(t)dt= t f^{-1}(t)-F( f^{-1}(t))+"c"`

We let `f^-1(x)=arctanx` and `f(x)=tanx`. Then `F(x)=inttanxdx=ln|secx|` and `F(f^-1(x))=ln|secarctanx|=lnsqrt(1+(tanarctanx)^2)=ln(sqrt(1+x^2))=1/2ln(1+x^2)`

So

`intarctanxdx=xarctanx-1/2ln(1+x^2)+"c"`