How do you find #int arctanx #?

3 Answers
Apr 12, 2018

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#

Apr 12, 2018

# 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#.#.

Apr 12, 2018

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"#